PLUGSINSIMPLY-CONNECTED4MANIFOLDSWITHBOUNDARIES By WeiFan ADISSERTATION Submittedto MichiganStateUniversity inpartialentoftherequirements forthedegreeof Mathematics-DoctorofPhilosophy 2015 ABSTRACT PLUGSINSIMPLY-CONNECTED4MANIFOLDSWITHBOUNDARIES By WeiFan In1986,S.BoyergeneralizedFreedman'sresulttosimply-connectedtopological4man- ifoldswithboundaries.Heprovedinmanycases,theintersectionformandtheboundary determinethehomeomorphismtypesofthe4manifolds.Inthisthesis,wewillstudysimply- connectedsmooth4manifoldswithboundariesbyusinghandlebodytechniques.Wewill showthat:theredoexistsimply-connectedsmooth4manifoldswiththesameintersection formandthesameboundarybutnothomeomorphictoeachother,andthecauseofthis phenomenonis\Plug". ACKNOWLEDGMENTS IwouldlikegratefullyandsincerelythankmyadvisorSelmanAkbulutforhisguidance, support,patienceandencouragementduringmyPhDprogramatMichiganStateUniversity. Hisuniqueperspectiveonresearchandsharpinsightonalmostanyissuehavebeenan invaluableresourcetomeinmystudies. IwouldalsoliketothankMattHeddenandKalfagianniformanyenlightening conversationsonmanytopics.IamindebtedLukeWilliamsforuncountablymanyinspira- tionaldiscussionsandhiscontinuedsupportforLatex.Iamthankfultomyfellowgraduate studentsatMSU.Ihavelearnedagreatdealfrommanyofyou.Iappreciatetheatmosphere ofmathematicallearningduringmyyearshere. Lastbutnotleast,IamgratefultomydadZhenfengFan,mymomRongmeiYangand myCaijinHuangforalwayshavingfaithinme. iii TABLEOFCONTENTS LISTOFFIGURES ................................... v Chapter1Introduction ................................ 1 Chapter2BackgroundandKnownResults ................... 6 2.1Whitney'sTrickandFingerMove........................6 2.1.1Whitney'strick..............................6 2.1.2Fingermove................................10 2.1.3Cassonhandles..............................12 2.2UsefulResultsinHandlebodyTheory......................14 2.2.1Pluming2-handles............................14 2.2.2DoingSurgery...............................17 2.2.3Eliminating1-handles..........................18 2.3SpinStructures..................................20 Chapter3CorksandPlugs ............................. 23 3.1Corks.......................................23 3.2Boyer'stheorem..................................25 3.3Plugs........................................31 Chapter4AnExample ................................ 47 4.1Anirreducible3manifold............................47 4.2Finding H + ( M )by Out ( ˇ 1 ( M )).........................55 BIBLIOGRAPHY .................................... 80 iv LISTOFFIGURES Figure2.1:Whitney'sdisk..............................6 Figure2.2:Whitney'smove.............................7 Figure2.3:Interiortwisting.............................8 Figure2.4:Boundarytwisting............................9 Figure2.5:Pushinteriorself-intersections...................9 Figure2.6:Fingermovechangesthefundmentalgroupofthecomplement...10 Figure2.7:Fingermove...............................11 Figure2.8:Eliminatingtheintersectionpointbyatransversesphere......12 Figure2.9:HandlebodypictureofWhitmey'smove................12 Figure2.10:Cassonhandle..............................13 Figure2.11:Handledecompositionof( D;@D )...................15 Figure2.12:Pluming2-handles............................15 Figure2.13:Pluming2-handlesin\dot"notation..................16 Figure2.14:Twospheresintersectinggeometrically3times,algebraicallyonce.16 Figure2.15:Atail................................17 Figure3.1:Anexampleofbeinghomeomorphicbutnotobtained bydoingcorktwist...........................24 Figure3.2:Apotentialexampleofbeinghomeomorphicbutcannotbestablized byconnectingsumwith S 2 S 2 bydoingcorktwist.........25 Figure3.3:Handledecompositionofthecobordism................33 Figure3.4:Finerhandledecompositionofthecobordism.............36 Figure3.5:Theneighborhoodof ' ( 0 ), 0 0 , 0 0 in X 2 ]S 2 e S 2 ...........38 Figure3.6:FindingaWhitney'sdisk........................39 v Figure3.7:Theneighborhoodof ' ( 0 ), 0 0 , 0 0 in X 2 ]S 2 e S 2 afterWhitney'smove40 Figure3.8:Glugtwist................................43 Figure3.9:Theneighborhoodof ' ( 0 ), 0 0 , 0 0 in X 2 ]S 2 e S 2 ...........44 Figure3.10:Plugtwist................................45 Figure3.11:Ageneralplugtwist...........................46 Figure3.12:Corktwiststogetherwithaplugtwist.................46 Figure4.1:Asimply-connected4-manifold.....................49 Figure4.2:DoingGlucktwistgivesanon-homeomorphicsimply-connected4- manifold.................................50 Figure4.3:Findingthefundmentalgroupoftheboundary............51 Figure4.4:Move1.1.................................61 Figure4.5:Move1.2.................................62 Figure4.6:Move2..................................64 Figure4.7:Move3..................................65 Figure4.8:Move4..................................67 Figure4.9:Move5..................................69 Figure4.10:Move6..................................71 Figure4.11:Turningupside-down..........................77 Figure4.12:Flipover.................................78 Figure4.13:DoingGlucktwistoneither S , T orboth S and T ..........79 vi Chapter1 Introduction Thisthesiswillfocusonthestudyofhomeomorphismtypeofsimply-connectedsmooth4- manifoldswithboundaries.Oneofthemainyinstudying4-manifoldtheoryarises fromthefailureofh-cobordismtheorem.In1960,S.Smale[41]provedfordimensionhigher than4,iftwosimply-connectedsmoothmanifoldsareh-cobordant,thentheyare phictoeachother. Unfortunately,indimension4,oneofthekeyingredientofSmale'sproof:theWhit- ney'strickdoesnotwork.In1973,A.Cassonlookedintothe4-dimensionalh-cobordism model,andintroducedanewidea\self-pluminghandle"whichiscalled\CassonHandle" nowadays(Cassonhimselfcalledithandle").In1982,M.Freedman[24]proved thattheCassonhandlesaretopologicallystandardhandles,andhencetheh-cobordismbe- tweentwosimply-connectedsmooth4-manifoldsactuallyinducesahomeomorphism.By combiningwithWall'stheorem[44],[45]whichclaimsthatanytwoclosedsimply-connected 4-manifoldswithisomorphicintersectionformsareh-cobordant,hederivedthefollowing celebratedtheorem: Theorem1.1. (Freedman,[24]) Foreveryunimodularsymmetricbilinearform Q there existsasimply-connected,closed,topological4-manifold X suchthat Q X ˘ = Q .If Q iseven, thismanifoldisunique(uptohomeomorphism).If Q isodd,thereareexactlytwoent homeomorphismtypesofmanifoldswiththegivenintersectionform.Atmostoneofthese 1 homeomorphismtypescarriesasmoothstructure. Asaconsequence,thefamousPoincareConjecturewasproved(the4-dimensionaltopo- logicalcase) Corollary1.1. (Freedman,[24]) Ifatopological4-manifold X ishomotopyequivalentto S 4 ,then X ishomeomorphicto S 4 . Freedman'sTheoremsaysthehomeomorphismtypeofclosed,simply-connected,smooth 4-manifoldsarecompletelybytheirintersectionforms.However,theclasscation ofthetypeofsimply-connected,smooth4-manifoldsisstillamysteryand farfromachieving. Indimension4,homeomorphismtypeistfromtype;actually, dimension4isthelowestdimensionthatsuchphenomenahappens.Manifoldswhichare homeomorphictoeachotherbutnotphictoeachotherareusuallycalled\exotic manifolds".Inthefollowingdecades,peopleinventedmanyntmethodstoproduce exotic4manifolds:LogrithmicTransformation,RationalBlow-down[21],Fintushel-Stern KnotSurgery[22],etc.\CorkTwist"[9]isoneofthemethodsthatattractspeople'sattention inrecentyears.Corksarecontractiblesubmanifolds;corktwistmeansthereisaninvolution mapontheboundaryofthecork;wecutthecorkandglueitbackbytheinvolution map.ManyCorksarefairlysimple4-manifolds,whilebyoperatingtwistsonthem,abundant illuminatingandinterestingexotic4-manifoldswereconstructed[10],[11],[12],[13]. In1996[32],[18],itwasprovedthatanytwosimply-connectedsmooth4-manifoldswhich arehomeomorphicbysomecorktwists.Indeed,thefailureofsmoothh-cobordism theoremindimension4isduetothecorktwists.So,corktwistplaysafundamentaland importantroleinchangingthetype(smoothstructure)ofthe4-manifolds. 2 Freedman'stheoremisalsotrueforsimply-connected4-manifoldswhoseboundariesare homology3spheres,sinceinthiscase, Q X isstillularandhenceWall'stheoremstillholds. In1986,S.Boyer[15]studiedthehomeomorphismtypeofsimply-connected,topological 4-manifoldswhoseboundariesarearbitraryclosed,oriented,connected3-manifolds.He gaveanecessaryandtconditionunderwhichagivenhomeomorphismbetweenthe boundariesofthetwo4-manifoldscanbeextendedintotheinterior.Hisresultcanbe consideredasageneralizationofFreedman'sTheoreminthefollowingway: Theorem1.2. (Boyer,[15]) Let M denoteaclosed,oriented,connected3-manifold; ( Z n ;Q ) denoteabilinearformspace.If Q presents 1 H ( M ) ,wewilldenotethembyapair ( Q;M ) . Then: (i)if H 1 ( M; Q ) ˘ = 0 ,thereareatmost2homeomorphismtypesofsimply-connectedtopolog- ical4-manifolds ( X;M ) whoseintersectionformis Q X suchthat ( Q X ;M ) isisomorphic 2 to ( Q;M ) .Atmostoneofthesehomeomorphismtypescarriesasmoothstructure. (ii)if Q isodd,thereareatmost2homeomorphismtypesofsimply-connectedtopologi- cal4-manifolds ( X;M ) whoseintersectionformis Q X suchthat ( Q X ;M ) isisomorphicto ( Q;M ) .Atmostoneofthesehomeomorphismtypescarriesasmoothstructure. (iii)if Q iseven,thereareatmostSpin ( M ) = H + ( M ) manyenthomeomorphismtypes ofsimply-connectedtopological4-manifolds ( X;M ) whoseintersectionformis Q X suchthat ( Q X ;M ) isisomorphicto ( Q;M ) . Moreover,If H 1 ( M ) isfree,then: (i)if Q isodd,thereareexactly2homeomorphismtypesofsimply-connectedtopological 4-manifoldswhichareisomorphicto ( Q;M ) andtheybytheirKirby-Siebenmannin- variant. (ii)if Q iseven,thereareexactlySpin ( M ) = H + ( M ) manyhomeomorphismtypesofsimply- 3 connectedtopological4-manifoldswhichareisomorphicto ( Q;M ) . AlthoughBoyerworkswithinthetopologicalcategory,histheoremhasaveryniceim- plicationinthesmoothcase: Corollary1.2. twosimply-connectedoddsmooth4-manifoldswhichhaveisomorphic ( Q;M ) arehomeomorphic. Nowonemaywonderwhathappenstotheevensmooth4manifolds.Boyergaveagen- eralalgorithmonconstructingSpin( M ) = H + ( M )manynon-homeomorphicsimply-connected eventopological4-manifoldswhichhaveisomorphic( Q;M ),when H 1 ( M )isfree.However, hisconstructiondoesnotworkinthesmoothcase,sincehisproofusesFreedman'sresult [24]: Everyhomology3-sphereboundsacontractibletopological4-manifold whichfailsinthe smoothcase. Whetherthereexistnon-homeomorphicsimply-connectedevensmooth4-manifoldswhich haveisomorphic( Q;M )?Inthelastchapter,wewillgiveapositiveanswertothisquestion bypresentingaconcreteexample.Wewillhuntthroughallpossibleorientationpreserving self-homeomorphismsofahakenmanifold M . M boundstwosimply-connectedsmooth4- manifolds X 1 and X 2 ,whichinducespinstructures s 1 and s 2 respectivelyon M .Thenwe willeasilyseethatnoneoftheseself-homeomorphismsinterchangethetwospinstructures. So X 1 and X 2 mustbenon-homeomorphictoeachother. Inchapter3,wewillprovethatif( X 1 ;M 1 )and( X 2 ;M 2 )aretwosimply-connectedeven smooth4-manifoldssuchthat( Q X 1 ;M 1 )isisomorphicto( Q X 2 ;M 2 ),thenthehomeomor- phismtypesof( X 1 ;M 1 )and( X 1 ;M 2 )byasingle\PlugTwist".\PlugTwist"isa similaroperationasCorkTwist.Itwasandstudiedin[9].Itcomesnaturallyfrom manyexamples.Forinstance,GluckTwist[25]canbeconsideredasasimplestPlugTwist. 4 ThekeybetweenaCorkandaPlugisthataPlugcontainsaframing1(or-1)2- handle:twistingaroundthis2-handlemaychangethehomeomorphismtype;whilesincethe boundaryofacorkisahomology3-sphere,corktwistingneverchangesthehomeomorphism type. BycombiningwiththetheoremofCork,wewoulddrawthefollowingconclusion: Iftwosimply-connectedsmooth4-manifoldshaveisomorphic( Q;M ),thentheplugtwistis responsiblefortheoftheirhomeomorphismtypes;corktwistsisresponsiblefor theoftheirtypes(smoothstructures). Inchapter2,wewillintroducethenecessaryterminologiesandbackground. 5 Chapter2 BackgroundandKnownResults 2.1Whitney'sTrickandFingerMove 2.1.1Whitney'strick Whitney'strickisamethodforremovingapairofdoublepointsofoppositesign(each intersectionpointhasasignfromcomparingorientations);itworksperfectlyindimension > 5.Sinceitthisthesis,wewillonlyconcentrateon4-dimensionalcase,letusconsiderthe followingWhitney'strickmodelindimension4. Figure2.1:Whitney'sdisk Suppose A and B aretwoembeddedsurfacesin X 4 whichintersectintwopointswithop- positesign(bydimensioncounting,wemayassumeallintersectionsaretransverse).Choose apaththatlinksthetwointersectionpointsinside A andchooseanotherpathlinkingthetwo 6 intersectionpointsin B .Theunionofthetwopathsformacircle,whichiscalled Whitney's circle ;wealsoassumethatthiscircleboundsanembedded2-diskinthecomplementof A and B ,asshowninthefollowingpicture.Theinteriorofthisdiskiscalled Whitney'sdisk . TheWhitney'scircleconsistsoftwoparts: l 1 lieson A ; l 2 lieson B .Thereisa1-vector bundle 1 over l 1 whichistangentto A andnormalto W ;thereisalsoa1-vectorbundle 2 over l 2 whichisnormaltoboth W and B . 1 and 2 agreeattheircommonpoints.Denote theunionof 1 and 2 asa1-vectorbundleovertheWhitney'scircle.Theobstruction ofextending over W is ˇ 1 SO (2).Itiscalledthe framingobstruction .Forthismoment, letusassumetheframingobstructionistrivial(intuitively,thismeanwhenwepush theWhitney'sdiskparallelly,theWhitney'scirclesarenotlinkedontheboundaryofthe complement).Now,underalltheseassumptions,weseethatthereisanambientisotopy, called Whitney'smove ,supportedinaneighbourhoodof W whichmoves A toasurface A 0 disjointfrom B . A 0 canbedescribedasconstructedfrom A bycuttingoutaneighbourhood ofthearcon A ,glueingintwoparallelcopiesoftheWhitney'sdisk,andaparallelofa neighbourhoodofthearcon B . Figure2.2:Whitney'smove TomaketheWhitney'strickworkindimension4,wemade3assumptions: 7 1.theWhitney'scircleboundsadiskinthecomplementof A and B . 2.theWhitney'sdiskisembedded. 3.theframingobstructionistrivial. Ingeneralsituations,noneoftheseassumptionscanbeguaranteed.Itseemsthereare toomanyissues.However,wemayreducethenumberofbyusingthefollowing tricks[38],[27]: (i)Interiortwisting: Thisoperationcreatesaself-intersectionof W ,whilechangestheframingobstructionby 2. Figure2.3:Interiortwisting (ii)Boundarytwisting: Thisoperationcreatesanewintersectionpointbetween W and A ,whilechangestheframing obstructionby 1. 8 Figure2.4:Boundarytwisting (iii)Pushinteriorself-intersections: Thisoperationeliminatesoneself-intersectionof W ;createstwonewintersectionpoints between W and A .Itdoesnotchangetheframingobstruction. Figure2.5:Pushinteriorself-intersections 9 So,ifwecanaWhitney'sdisk W whichisdisjointform B ,evenifitisimmersed, theframingiswrong,wecanmodifyitby(ii)and(iii)suchthatthenewWhitney'sdiskis embedded,andtheframingobstructionbecomestrivial.Thepricetopayisweobtainmore intersectionpointsbetween W and A .NowwecanstilldotheWhitney'strickbydragging A alongthenewWhitney'sdisk.Theisthatwegetasurface A 0 whichisdisjointfrom B ,but A 0 isimmersed. 2.1.2Fingermove TouseWhitney'strickindimension4(toeliminateintersectionpointsbetween A and B withoutgettingself-intersections),weneedtoaWhitney'sdiskinthecomplementof A and B Thissuggestsusthecomplementof A and B betterbesimply-connected. Abrilliantideaofreducingthefundamentalgroupofthecomplementof A and B isdue toCasson:ImaginewepushourthroughAfollowingaloop inthecomplement,as showninthefollowing Figure2.6:Fingermovechangesthefundmentalgroupofthecomplement Thismethodiscalled Fingermove .Bydoingthemove,wecreatetwonewinter- sectionpointsbetween A and B ,soitcanberegardedas\inverseWhitney'strick",andwe willdenotethisWhitney'sdiskby W .Nowletusconsiderhowdoesthisprocessthe 10 fundamentalgroupofthecomplement: Aneighbourhoodof p islocallyto( R 4 ˘ = R 2 xy R 2 zt ; R 2 xy [ R 2 zt ).Thereis atorus T 2 = S 1 S 1 ontheboundaryofthecomplementof R 2 xy [ R 2 zt in R 4 whichlinks R 2 xy [ R 2 zt .( T 2 = f ( x;y;z;t ) 2 R 1 j x 2 + y 2 =1= z 2 + t 2 g ).Thistorusiscalled the linkingtorus .Itsfundamentalgroupisgeneratedexactlyby and .Sobyintroducing newintersectionpoints,weaddarelationinducedbythelinkingtorustothefundamental groupofthecomplement.Thusmovekillsthecommutator[ ; ]. Pushingtheinteriorself-intersectionscanbealsoviewedasanapplicationof move.Ingeneral,suppose A , B , C aresurfacesin X 4 ,partoftheboundaryof C lieson B , andthereisanembeddedarcfromanintersectionpoint A \ C tothisboundary,asshown. Thenwecanpush A C through B (alongthearc).Thisgivesasurface A 0 withonefewer intersectionwith C ,buttwonewintersectionswith B . Figure2.7:Fingermove If B isaspherewhoseframingis0(meansaparallelpushof B in X doesnotintersect B )andintersect C inexactlyonepoint.Thenalltheintersectionsbetween A and C can beremoved,bypushing A alongthearcin C andaddingparallelcopyof B to A .Inthis situation, B iscalleda transversesphere of C . 11 Figure2.8:Eliminatingtheintersectionpointbyatransversesphere 2.1.3Cassonhandles Nowletusassumethat A;B areembeddedsurfacesin X ,suchthat A \ B = f p;q g with oppositesignandthe X n A [ B issimply-connected,thenweareabletoaWhitney's disk W disjointfrom A and B .If W isembeddedsuchthattheframingobstructionistrivial (wemaythinkofitasthecoreofa0framing2-handleasinthefollowingpicture),thenby Whitney'strick(wemaythinkofitashandleslidesoverthis0-framing2-handle),wecan removethetwointersectionpointswithoppositesign. Figure2.9:HandlebodypictureofWhitmey'smove However,ifthe W isjustanimmerseddisk,itwoulda kinkyhandle ,i.e.,the coreofthis2-handlehasself-intersections.Inthiscase,wecanalwaysadjustitsframing 12 byinteriortwistingsuchthatafterintroducingmoreself-intersectionpoints,theframing obstructionistrivial(sinceforapairofintersectionpointsofoppositesign,theframing obstructionis 0mod(2)).Foreachself-intersectionpointof W ,choosealoopin W ,based attheself-intersection,leavingalongonebranchandreturningalongtheother.Wewant tochoosetheseloopspairwisedisjoint.Thereisnoframingissuefortheseloops,sincethey areentirelycontainedin W .Thelinkingtorus T 2 p at p intersects W atasinglepoint,soby M-Vsequence,wecaneasilycheckthat H 1 ( X n A [ B [ W )=0.Thus ˇ 1 ( X n A [ B [ W )is aperfectgroupandgeneratedbytheconjugatesofthemeridiancircleof W .Now,byusing moveson W (wepush W throughitself),wecankill ˇ 1 ( X n A [ B [ W ).Therefore, everyloopboundsadiskin X n A [ B [ W .Ifoneoftheseloopsactuallyboundsan embeddeddisk,thenthecorrespondingself-intersectionof W canbeeliminated,byhandle slideorWhitney'strick(wecancreateanewself-intersectionwithoppositesignnearthe boundaryofthisembeddeddisk([40],2.1)).Ingeneral,wecanmerelyanimmersed disk,withitsownself-intersections,whichcanbeviewedasakinkyhandleattachedtoa kinkyhandle.Weiteratetheprocess;addthirdstagekinkyhandlesandthenfourthstage kinkyhandles,andsoon.Wecarryouttheprocessformanysteps,thentakethe unionofallthesekinkyhandles.Theresultiscalled Cassonhandle . Figure2.10:Cassonhandle Fromtheconstruction,weseethattheCassonhandlesareverycomplicatedsubjects, 13 butthemiracleistheyareactuallytopologicallythesameasstandardhandles. Theorem2.1. (Casson) Let X 4 besimply-connectedandlet W 1 ;:::;W n besmoothlyim- mersedtransverse2-disksin X 4 withboundaries, @W 1 ;:::;@W n ,embeddeddisjointlyin @X 4 , and W i W j =0 for i 6 = j .Assumethatthereexist 1 ;:::; n 2 H 2 ( X ) suchthat i i iseven and W i j = ij .Thenthe W i 'scanberegularlyhomotoped(rel @ )tobedisjoint,andthen kinkyhandlesmaybeaddeddisjointlysoastobuildndisjoint,smoothlyembeddedCasson handleswith W 1 ;:::;W n astheirstages.Furthermore,theseCassonhandlessatisfy: Property1:EachCassonhandleisproperhomotopyequivalent,rel @ = S 1 int B 2 ,to B 2 int B 2 . Property2:EachCassonhandleisasmoothsubmanifoldof B 2 B 2 with S 1 int B 2 = @B 2 int B 2 . Theexistenceof i ;i =1 ;::;n intheabovetheoremistoguaranteethecomplementof W i ;i =1 ;:::;n in X 4 issimply-connectedafternecessarymoves. Theorem2.2. (Freedman,[24]) EachCassonhandleishomeomorphicto B 2 int B 2 ,rel @ . 2.2UsefulResultsinHandlebodyTheory 2.2.1Pluming2-handles Themainobjectiveofthissectionistothehandlebodypictureoftheneighbourhood oftwoimmersedspheres A , B in X 4 withintersectionpoints.Adetaileddescriptioncanbe foundin[26]. Attheneighbourhoodofeachintersectionpoint,theneighbourhoodof A isphic to D 2 D 2 ,callit A 0 ;theneighbourhoodof B isto D 2 D 2 ,callit B 0 .The 14 neighbourhoodof A [ B canbethoughtasplumbingtwonormalbundles D 2 D 2 and D 2 D 2 .Wecanplumbthemasfollows:Staringwiththerelativehandledecompositionof thecore( D;@D )of A 0 withasinglehandle,introduceacancelling0-and1-handle. Figure2.11:Handledecompositionof( D;@D ) Weidentifythe0-handlewithacocoreof B 0 (realizedasadiskin X boundedbya meridianoftheattachingcircleof B 0 ).Thenweattach A 0 byattachingthe1-handleand 2-handle. Figure2.12:Pluming2-handles Ifweswitchto\dot"notation,itisshownasinthefollowing 15 Figure2.13:Pluming2-handlesin\dot"notation Example1:Aframing2sphereintersectswithaframing3spheregeometrically3times, algebraicallyonce. Figure2.14:Twospheresintersectinggeometrically3times,algebraicallyonce Notethatattheclasp,the1-handlelinksthetwo0-handlescomingfromeachhandle- bodypictureof A and B ,soitiscancelled.Aftertheclasp,wewillassociatea1-handle toeachclasp. Example2:Aframing0spherewithaself-intersection.(called a ) 16 Figure2.15:Atail 2.2.2DoingSurgery 1. Let ' : S k ! M n ( 1 k n )beanembeddingofa k -sphereinan n - manifold,withanormalframing f on ' ( S k )(whichweassumeliesinint M ).Thenthepair ( ';f )determinesanembedding^ ' : S k D n k ! M (uniqueuptoisotopy),and surgery on( ';f )istheprocedureofremoving^ ' ( S k int D n k )andreplacingitby D k +1 S n k 1 , withgluingmap^ ' j S k S n k 1 . Attachingahandleto( X;@ X )hastheofsurgeryon @ + X ,andconverselyany surgeryonaclosedmanifold M isrealizedas @ + ( I M [ h )where h isattachedby( ';f ). Surgeryon M producesamanifoldwithacanonicalembeddingof D k +1 S n k 1 .Ifwe surgeronthisframed S n k 1 ,werecover M .Thiscorrespondstoturningtherelative handlebody I M [ h upsidedown.Wecallthisprocedure reversethesurgery . Now,letusconsidertheparticularcasewhen n =4.If k =1,wecutaframed embedding S 1 D 3 from M ,andgluein S 2 D 2 .Asexplainedabove,thisisthesame asattachinga5-dimensional2-handleto I M .Ifwestartbuildingthehandlebodyof M from S 1 D 3 ,itcanbeviewedasa4-dimensional1-handle;Replacingitby S 2 D 2 by usingthecanonicalframingisequivalenttoswitchingthe1-handletoa0framing2-handle. Clearly,switchingthe0framing2-handletoa1-handle(acirclewithadot)isthereverse 17 surgery. Proposition2.1. Suppose X isasimply-connected4-manifold,thenany1-surgery( k =1 ) on X yieldseither X]S 2 S 2 or X]S 2 ~ S 2 . Proof. Supposewedo1-surgeryonacircle C ˆ X .Write X = X]S 4 ,andlet C 0 ˆ X]S 4 bethecircle @D 2 0 ˆ @ ( D 2 D 3 )= S 4 . ˇ 1 ( X )=1,so C mustbehomotpicto C 0 and henceisotopicto C 0 in X ,asdim X =4.Therefore,doing1-surgeryon C 0 wouldgiveusthe sameresult.Since ˇ 1 ( SO (2)) ˘ = Z 2 ,wecanview S 4 asa1-handlelinkingwitha0framing 2-handleonceora1-handlelinkingwitha1framing2-handleonce.1-surgeryswitchesthe 1-handletoa0framing2-handle,whichyields X]S 2 S 2 or X]S 2 ~ S 2 respectively. 2.2.3Eliminating1-handles Let W beanorientedcompact5-manifoldwithboundary @W decomposedasadisjoint union @ + W ` @ W oftwocompactsubmanifolds(eitherofwhichmaybeempty).Let @ W = X , @ W = X + ,wesay( W;X )isarelativehandlebodyif W isobtainedfrom I X byattaching1,2,3,4-handles(if X isempty,westartfroma0-handle;if X + is empty,weendupwitha5-handle). Theorem2.3. If W issimply-connectedand X , X + areconnected,thenwecanmodify thehandledecompositionof W suchthatitcontainsno1-handlesand4-handles. Proof. Foreach1-handle,wecanintroduceacancellingpairof2and3-handles.Wehope eachnewintroduced2-handlewouldcancelwitha1-handle.Thisistrueiftheattaching circle K ofthe2-handleisisotopictothecore K 0 ofthe1-handlein @ + W 2 ,where W 2 is the2skeletonof W ,i.e.,itistheunionof I X andallthe1-handlesand2-handles. Sincedim @ + W 2 =4,itisenoughtoshowtheyarehomotopicin @ + W 2 . ˇ 1 ( W )=1implies 18 that W 2 isalsosimply-connected.Therefore,thereisahomotopybetween K and K 0 in W 2 .Asdim W 2 =5,dimensionofthishomotopyis2,anddimensionsofthecoresofthe 1-and2-handlesare1and2respectively,wecanmakethehomotopytobedisjointfrom 1-and2-handlesbysmallperturbationin W 2 .Therefore,thishomotopycanbepushed into @ + W 2 .Thus,allthe1-handlesarecancelledwiththenewintroduced2-handles.The newintroduced3-handlesareleft,sowecanthinkthisprocessas\trading"1-handlesfor 3-handles.Byturning W upside-down,wecantradeeach4-handlefora2-handle.Inthe end,wegetahandledecompositionof W involvingno1-handlesand4-handles. Fromtheproof,wecaneasilyseethatthesameresultistruefordimension > 5. Theproofclearlyfailsindimension4.Indeed,whetherasimply-connectedclosed4- manifoldalwaysadmitsahandledecompositioninvolvingno1-handleisanopenquestion. However,wecanusethesameideaasintheproofoflasttheoremtogetaweakerresultin dimension4. Theorem2.4. Supposethat ( X;@ X ) isa4-dimensionalrelativehandlebodywith @ X connectedand ˇ 1 ( X )=1 .Forevery1-handle h ofthisdecomposition,onecanintroducea cancelling2-3handlepairsuchthatthe2-handlecancels h algebraically(butnotnecessarily geometrically).Infact,onecanarrangetheattachingcirclesofthenew2-handlestorepre- sentthecanonicalbasisforthefreefactorof ˇ 1 ( X 1 ) determinedbythe1-handles(suitably attachedtothebasepoint). Proof. Ifwegothroughtheproofoflasttheoremfordimension4,wewillrealizethesame techniquefailsintwoplaces:1.Indimension4,homotopydoesnotimplyisotopy;2.We cannotassumethehomotopymissesthecoresofthe2-handles. WedealwiththesecondproblemUsethesamenotationasintheproofoflast 19 theorem: K 0 denotethecoreofa1-handle h ; K denotetheattachingcircleofthecancelling 2-handle; H denotethehomotopybetween K and K 0 .Suppose H intersectsthecoreofa 2-handleatapoint P ,thenwecanformtheband-sumoftheboundarycircleofasmall diskaround P in H withthecircle K 0 usingabandcontainedin H ;theresultingcirclewill bedenotedby K 1 .Now, K 1 ishomotopicto K 0 andthishomotopycanbeassumedtobe disjointformthe2-handlesandhencecanbepushedinto @ + W 2 .Denotetheimageof K 1 on @ + W 2 by K 2 .Notethatthebandmightrunover1-handlesand2-handles,so, K 2 might runover1-handlesand2-handlesaswell,butitwillrunover h algebraicallyonce,other 1-handlesalgebraically0times. Nowwedealwiththeproblem.Dim @ + W 2 =3,thehomotopyin @ + W 2 failsto beanisotopyatmanytimeswhentheknotcrossesthroughitself.Eachcrossing changecanberealizedbyband-summing K 2 withameridian K 2 alongasuitableband in @ + W 2 .Noteagain,thisbandmightrunover1-handlesand2-handles,butitdoesnot changethealgebraiclinkingnumberbetween K 2 andthecoreofeach1-handle.Let K 3 betheknotobtainedfrom K 2 bysuchcrossingchanges. K 3 runover h algebraicallyonce, other1-handlesalgebraically0timesanditisisotopicto K in @ + W 2 .So,the2-handle K algebraicallycancels h . 2.3SpinStructures Let X denotean-manifold n 3, T X denoteitstangentbundle.ThesecondStiefel-Whitney class ! 2 ( T X )measurestheobstructiontotrivializing T X overthe2-skeletonof X . When n =4,Wu'sformulasays: Forallorientedsurfaces S embeddedin X , ! 2 ( T X ) S = S S (mod2). 20 AniceconsequenceofWu'sformulais: If ! 2 ( T X )=0 ,thentheintersectionformof X iseven. Byuniversalcottheorem,theconverseistruewhenever H 1 ( X ; Z )hasno2-torsion. A spinstructure on X isachoiceoftrivializationof T X over1-skeletonthatcanbe extendedoverthe2-skeleton,considereduptohomotopies.Amanifoldendowedwithaspin structureiscalleda spinmanifold . Note:When n 2,weaspinstructureon X asatrivializationof T X overthe 1-skeletonsuchthat T X ˘ 3 n canbeextendedover2-skeleton,where ˘ 3 n isatrivialized bundle. ByWu'sformula, Any4-manifoldwithout2-torsion,forexamplesimply-connected,ad- mitsspinstructuresifandonlyifitsintersectionformiseven. Let s beaspinstructureon X n ,givenbyatrivializationof T X overthe1-skeletonof X (forsometriangulationof X ).Forany 2 H 1 ( X ; Z 2 ), isisomorphictoamap f : X ! RP 1 .Wehomotope f intoaskeleton,say RP N ,thenthepreimageof RP N 1 isa n 1submanifold Y of X ,representing .Nowwechangethetrivializationof s over every1-cellbya2 ˇ twisteachtimethe1-cellintersects Y .Ifa1-cellboundsa2-cell,then this1-cellmustintersect Y evennumberoftimes,andthuswechangethetrivializationover this1-cellbyevennumberof2 ˇ twists.Since ˇ 1 ( SO (4)) ˘ = Z 2 ,thenewtrivializationcan alsobeextendedoverthe2-skeleton.Therefore,wegetanewspinstructureon X ;denoteit by s .Since s and arearbitrarilychosen,thiscanbeviewedasanactionof H 1 ( X ; Z 2 ) onthesetofallpinstructuresof X .Thisactionisfreeandtransitive.Therefore,after aspinstructureon X ,theactionestablishesabijectivecorrespondencebetweenthe elementsof H 1 ( X ; Z 2 )andthesetofallspinstructureson X . Forexample, H 1 ( S 1 ; Z 2 )= Z 2 ,sotherearetwospinstructureson S 1 .Onecanbe 21 extendedintotheinterior;onecannot.Thelatteroneisalsocalled theLiegroupspin structure on S 1 . The3-torus T 3 = S 1 S 1 S 1 has8tspinstructuressince H 1 ( T 3 ; Z 2 )= Z 3 2 , whichcanbealsoviewedastheproductsofspinstructureson S 1 . Foraclosed,spin3-manifold( M;s ),the Rohlininvariant ( M;s ) 2 Z 16 isthesignature ˙ ( X )reducedmodulo16,where X isanysmooth,compact,spin4-manifoldwithspin boundary( M;s ) Thisisawinvariantbecauseif X and Y aretwosmoothcompactspin4- manifoldswhichinducethesamespinboundary( M;s ),then X [ M Y isasmoothclosed spin4-manifold.TherestoftheargumentisduetoRohlin'sTheorem: Theorem2.5. (Rohlin,[37]) If X isasmooth,closed,spin4-manifold,then ˙ ( X ) 0 (mod16). 22 Chapter3 CorksandPlugs 3.1Corks Inthischapterandthenextchapter,whenwetalkaboutamanifold,wealwaysassumeitis compactandoriented.\ ˇ "standsfor\ ' "standsforhomotopicequivalent. Theh-CobordismTheoremisoneofthemostimportanttheoreminmoderntopology. Theorem3.1. (Smale,[41]) If W isanh-cobordismbetweenthesimply-connected n -dimen- sionalsmoothmanifolds X 1 and X 2 ,and n > 5 ,then W isomorphictotheproduct I X 1 .Inparticular, X 1 isomorphicto X 2 . However,thistheoremfailsindimension4,thecounterexamplewasbroughttolight byS.K.Donaldson[19],andmanyothersfollowed.Thefollowingtheoremwhichwasproved independentlybyMatveyev[32]andCurtis-Freedman-Hsiang-Stong[18]tellsusthatthe failureofh-cobordismtheoremcanbelocalizedonacontractiblepiece. Theorem3.2. ([32],[18]) X 1 and X 2 aresimply-connectedsmooth4-manifolds(notneces- sarilyclosed).If W isan(relative)h-cobodismbetween X 1 and X 2 ,thenthereisasubcobor- dism V ˆ W betweensubmanifolds C i ˆ X i ( i =1 ; 2) ,suchthat W int V istheproduct cobordism,and V , C i arecontractible. Thistheoremsaysif X 1 ish-cobordantto X 2 ,then X 1 , X 2 canbedecomposedas X 1 = X 0 [ @ C 1 , X 2 = X 0 [ @ C 2 ,where C 1 , C 2 arecontractiblemanifolds.Inmanycases, 23 C 1 ˇ C 2 .Actually,in[32],Matveyevprovedthat C 1 [ @ C 1 ˇ S 4 ; C 1 [ @ C 2 ˇ S 4 and sothat X 1 ˇ ( X 0 \C 1 ) [ @ ( C 1 \C 2 )and X 2 ˇ ( X 0 \C 1 ) [ @ ( C 2 \C 1 ).Thusiflet C 1 \C 2 and C 2 \C 1 playtheroleof C 1 and C 2 ,onecanalwaysassumethat C 1 ˇ C 2 .ByFreedman andQuinn'stheorem[24],[36], X 1 and X 2 areh-cobordantimpliestheyarehomeomorphic. Therefore,wecanconcludeanyexoticsmoothstructureofasimply-connected4-manifold arisesfromcuttingacontractiblesubmanifoldandgluingitbackbyanon-trivialmap ˝ . FromMatveyev'sproof,wecaneasilyseethatthisnon-trivialmap ˝ isaninvolution,i.e., ˝ ˝ = id .Furthermore,in[1],AkbulutandMatveyevprovedthatonecanalwaysmake C i stein. Let C beacontractiblestein4-manifoldwithboundaryand ˝ : C ! C aninvolutionon theboundary.Wecall( C;˝ )a Cork if ˝ extendstoaself-homeomorphismof C ,butcannot extendtoanyof C .Theprocedureofcutting C andgluingitback by ˝ iscalled CorkTwist . TheexampleofcorktwistchangingthesmoothstructureisduetoS.Akbulut[3]: 6ˇ Figure3.1:Anexampleofbeinghomeomorphicbutnotobtainedbydoing corktwist ManyotherinterestingexoticmanifoldswerethenconstructedusingcorktwistsbyAk- bulutandYasui[9],[10],[11],[12],[13]. Notethatmostoftheseexampleswereconstructedbya\simple"corktwist,whichmeans 24 weonlyexchange\dot"and\0"once.Therefore,connectingsumwith S 2 S 2 stabilizes thesemanifolds,i.e., X 1 ]S 2 S 2 ˇ X 2 ]S 2 S 2 .Easytoseethatif X 1 and X 2 by n disjointsimplecorktwist,itisstilltruethat X 1 ]S 2 S 2 ˇ X 2 ]S 2 S 2 .Sotoconstruct exoticmanifoldswhichcannotbestabilizedbyasingle S 2 S 2 ,onehastouse\linked corks".Thefollowingisapotentialexample: Figure3.2:Apotentialexampleofbeinghomeomorphicbutcannotbestablizedbycon- nectingsumwith S 2 S 2 bydoingcorktwist 3.2Boyer'stheorem Inthissection,wewillinterpretBoyer'swork,whichwillbeneededinthenextsection. Let M denoteaclosed,oriented,connected3-manifold.Abilinearformspace( Z n ;Q )issaid to present H ( M )ifthereisanexactsequence 0 ! H 2 ( M ) h ! Z n ad( Q ) ! [ Z n ] @ ! H 1 ( M ) ! 0 suchthat (i)ifad( Q )( ˘ i )= m i i ( i =1 ; 2)where m 1 m 2 6 =0,then 25 l M ( @ 1 ;@ 2 ) 1 m 1 m 2 Q ( ˘ 1 ;˘ 2 ); (ii)if 2 H 2 ( M )and 2 [ Z n ] ,then @ ( ) = ( h ( )). Itiswellknownthat,if M boundsatopological4-manifold X ,( H 2 ( X ) ;Q X )presents H ( M ),where Q X istheintersectionformof X .If Q presents M ,wedenotethemby ( Q;M ). Giventwosimply-connected4-manifolds( X 1 ;M 1 )and( X 2 ;M 2 ),wesay( Q X 1 ;M 1 )is isomorphic(through ;f ) ) to( Q X 2 ;M 2 )ifthereisahomeomorphism f : M 1 ! M 2 andan isomorphism: H 2 ( X 1 ) ! H 2 ( X 2 )preservingtheintersectionformsuchthatthefollowing diagramcommutes: 0 ! H 2 ( M 1 ) ! H 2 ( X 1 ) ! H 2 ( X 1 ;M 1 ) ! H 1 ( M 1 ) ! 0 f ? ? y ? ? y x ? ? f ? ? y ( ) 0 ! H 2 ( M 2 ) ! H 2 ( X 2 ) ! H 2 ( X 2 ;M 2 ) ! H 1 ( M 2 ) ! 0 Note: istheadjointofwithrespecttotheidenof H 2 ( X i ;M i )with Hom( H 2 ( X i ); Z );If F : X 1 ! X 2 isahomeomorphismwhichinduces f onthebound- ary, F : H 2 ( X 1 ;M 1 ) ! H 2 ( X 2 ;M 2 )isnotthesameastheinverseofadjointof F : H 2 ( X 1 ) ! H 2 ( X 2 )ingeneral,but(*)iscommutativeduetoLefschetzduality.Therefore, (*)isanecessaryconditionfor X 1 , X 2 tobehomeomorphic. ThefollowingtheoremisduetoBoyer: Theorem3.3. (Boyer,[15])( X 1 ;M 1 ) , ( X 2 ;M 2 ) aresimply-connectedtopological4-manifolds withboundaries.If f : M 1 ! M 2 isanorientationpreservinghomeomorphismand : H 2 ( X 1 ) ! H 2 ( X 2 ) isanisomorphismwhichpreservestheintersectionform,suchthat ( Q X 1 ;M 1 ) isisomorphicto ( Q X 2 ;M 2 ) through ;f ) ,thenfcanbeextendedtoahomeo- morphism F : X 1 ! X 2 if: (i)when H 1 ( M i ; Q ) ˘ = 0( i =1 ; 2) and X 1 ) X 2 ) (mod2). 26 (ii)when X i ( i =1 ; 2) areevenmanifolds,and X = X 1 [ f X 2 isalsoanevenmanifold. (iii)when X i ( i =1 ; 2) areoddmanifoldsand X 1 ) X 2 ) (mod2). Incase(i)and(ii), F ,theisomorphismon H 2 ( X 1 ) inducedby F agreeswith ;incase (iii) F maynotagreewith . Fromthistheorem,onecaneasilyderivethentheoremwestatedinchapter 1. Inthisthesis,weareonlyinterestedinsmooth4-manifolds,soletusassume X 1 and X 2 aresmoothanddroptheassumption X 1 ) X 2 )(mod2). Tomakehistheoremmoretransparent,letusintroducesomenotations:Asstatedin thetheorem,( X 1 ;M 1 ),( X 2 ;M 2 )aresimply-connectedsmooth4-manifoldswiththeclosed connectedboundaries. f : M 1 ! M 2 isanorientationpreservinghomeomorphism.: H 2 ( X 1 ) ! H 2 ( X 2 )isanisomorphismwhichpreservestheintersectionform,suchthat ( Q X 1 ;M 1 )isisomorphicto( Q X 2 ;M 2 )through ;f ).Denote X 1 [ f X 2 by X .If: A ! B isahomomorphismofabeliangroups.Let G denotethesubgroupof A B corresponding tothegraphof: G = f ( a; a )) j a 2 A g . k , h and @ 1 arebythefollowing diagram: H 2 ( M 1 ) H 2 ( M 1 ;M 1 )=0 ? ? y ? ? y H 2 ( M 1 ) L H 2 ( M 2 ) h 1 h 2 ! H 2 ( X 1 ) L H 2 ( X 2 ) j 1 j 2 ! H 2 ( X 1 ;M 1 ) L H 2 ( X 2 ;M 2 ) ? ? y 1+ f 1 ? ? y i = i 1 + i 2 H 2 ( M 1 ) h ! H 2 ( X ) k ! H 2 ( X;M 1 ) @ 1 + f 1 @ 2 ! H 1 ( M 1 ) ? ? y @ ? ? y H 1 ( M 1 ) H 1 ( M 1 ;M 1 )=0 ? ? y 0 Boyer'stheoremisessentiallyaconsequenceofthefollowingtheorem: Theorem3.4. (Boyer,[15]) Assumerankof H 2 ( M 1 )= k .Thereexists f 0 i g2 H 2 ( X ) 27 ( i =1 ; 2 ;:::;k ),suchthat @ ( 0 i )= i ,where f i g isabasisofthefreepartof H 1 ( M 1 ) ,and k ( 0 i ) 2 G ( ) . f extendstoahomeomorphism F : X 1 ! X 2 if 0 i 0 i isevenforall i =1 ; 2 ;:::;k . Becauseif H 1 ( M i ; Q ) ˘ = 0( i =1 ; 2),thefreepartof H 1 ( M 1 )istrivial, k =0,so 0 i 0 i =0 mod(2)isautomaticallytrue;if X iseven, 0 i 0 i mustbeevenforany i ;if X isodd, 0 i 0 i mightbeoddforsome i .Inthiscase,Boyershowedthatonecanalwaysanother 0 and f 0 i g2 H 2 ( X )suchthat @ ( 0 i )= i , k ( 0 i ) 2 G ( )and 0 i 0 i isevenforevery i . Wewillinvestigatethiscasecarefullyandouthowtoconstruct 0 frominthenext section.Actually,anappropriate 0 isthekeytoprovetheexistenceoftheplug. Forcompleteness,weoutlineBoyer'sproofoftheabovetheorem. Proof. Theproofconsistsoftwosteps. 1.Weconstructamaximalisotropicsubgroup J H 2 ( X )(\isotropic"meansfor 8 x;y 2 J , x y =0)suchthat(i) i G ( J ;(ii) @ ( J )= H 1 ( M 1 ). 2.WeapplyWall'smethod[45]:replacing W byanother5-manifold W 1 with @W 1 = X , W 1 w W n i =1 S 2 and J = ker ( H 2 ( X ) ! H 2 ( W 1 )).Thenbydiagramchasing,wewillshow that i : H 2 ( X i ) ! H 2 ( W 1 )( i =1 ; 2)areisomorphisms. Step1: J willbebuiltasthesumoftwoisotropicsubgroups J 1 and J 2 of H 2 ( X ) whichsatisfy(i) i ( G ( ˆ J 1 , @ ( J 1 )= T 1 ( M 1 )(torsionof H 1 ( M 1 ))andrank( J 1 )= rank( H 2 ( X 1 )) rank( H 1 ( M 1 ));(ii)thecomposition J 2 @ j J 2 ! H 1 ( M 1 ) ! H 1 ( M 1 ) =T 1 ( M 1 ) isanisomorphism;(iii) J 1 \ J 2 = f 0 g and J 1 J 2 = f 0 g . Assumingwehavefoundsuchsubgroups,welet J bethesmallestdirectsummandof H 2 ( X )containing J 1 + J 2 .Asrank( H 2 ( X ))=2rank( H 2 ( X 1 )), J isevidentlythedesired subgroupof H 2 ( X ). 28 Constructionof J 1 : Let J 1 bethesmallestdirectsummandof H 2 ( X )containing i ( G ( whichisisotropic in H 2 ( X )asisanisometry.Thus J 1 isalso. Nextweprove @ ( J 1 )= T 1 ( M 1 ).If 2 J 1 ,thereisaninteger m> 0suchthat 2 i ( G ( Butthen m@ ( )=0,as @ i =0.Hence @ ( ) 2 T 1 ( M 1 )whichshows @ ( J 1 ) T 1 ( M 1 ).Toderivetheoppositeinclusion,let 2 T 1 ( M 1 ). Claim3.1. Wecanaclass 2 H 2 ( X ) with @ ( )= and k ( ) 2 G ( ) . Proof. Fixany 2 H 2 ( X 2 ;M 2 )suchthat @ 2 ( )= f ( ).Nowas( @ 1 + f 1 @ 2 )( G ( ))= 0, G ( ) ˆ image( k ),sowemaychoosea 2 H 2 ( X )forwhich k ( )=( ( ) ; ). Further, @ ( )= f 1 @ 2 pr 2 k ( )= f 1 @ 2 ( )= bythechoiceof . Let k ( )=( ( ) ; ).Since 2 T 1 ( M 1 ),wemaydan m> 0and ˘ 2 H 2 ( X 1 )such that j 1 ( ˘ )= m ( ).Then k i ( ˘; ˘ ))= m ( ( ) ; )andthus i ( ˘; ( ˘ ))= + h ( )forsome 2 H 2 ( M 1 ). Claim3.2. Theclass isdivisibleby m in H 2 ( M 1 ) Proof. Ittoshowthat 0(mod m )foreach 2 H 1 ( M 1 ).Butfromthe propertiesof J 2 ,foranysuch thereissome 0 2 J 2 with @ ( 0 ) 2 T 1 ( M 1 ).Then = h ( ) 0 = i ( ˘; ( ˘ )) 0 0 = 0 . Nowsince + h ( =m ) 2 J 1 , = @ ( )= @ ( + h ( =m )) 2 @ ( J 1 )andas waschosen arbitrarily,weconclude @ ( J 1 )= T 1 ( M 1 ). Finally,tocalculaterank( J 1 ),note ker ( i )= f ( h 1 ( ) ; h 2 f ( ) j 2 H 2 ( M 1 )) g G ( Thus, rank( J 1 )=rank( i ( G ( =rank( G ( rank( ker ( i ))=rank H 2 ( X 1 ) rank( H 1 ( M 1 )). 29 Constructionof J 2 : Set F 1 ( M 1 )= H 1 ( M 1 ) =T 1 ( M 1 )andchoose 1 ; 2 ;:::; k 2 H 1 ( M 1 )whichprojectsto abasisofthisgroup.ByClaim1,thereareclasses 0 1 ; 0 2 ;:::; 0 k 2 H 2 ( X )suchthat(i) @ ( 0 i )= i ; 1 i k ,(ii) k ( 0 i ) 2 G ( ) ; 1 i k . Let 1 ; 2 ;:::; k 2 H 2 ( M 1 )bethebasisdualto 1 ;:::; k .Thatis i j = ij .Set i = h ( i )(1 i k )andnotethat(i) i j =0 ; 1 i;j k ;(ii) i 0 j = ij ; 1 i;j k . 00 i = 0 i P k j = i +1 ( 0 i 0 j ) j ; 1 i k ,andobservethat k ( 00 i )= k ( 0 i ) 2 G ( ). Thusforeach i , 00 i 00 i = 0 i 0 i 0(mod2)as 0 i 0 i iseven.Thuswemayform i = 00 i 1 2 ( 00 i 00 i ) i 2 H 2 ( X ) ; 1 i k . Nowitcanbecheckedthat i j =0(1 i;j m ), @ ( i )= i (1 i k ),and k ( i ) 2 G ( ).Thusifweset J 2 = Span ( 1 ; 2 ;:::; k ) H 2 ( X ),(i) J 2 isisotropic; (ii)thecomposition J 2 @ j J 2 ! H 1 ( M 1 ) ! F 1 ( M 1 )isanisomorphism;(iii) k ( J 2 ) G ( ). Thus J 2 thedesiredproperties. Observethatunderthecomposition H 2 ( X ) @ ! H 1 ( M 1 ) ! F 1 ( M 1 ), J 1 mapstozerowhile J 2 mapsmonomorphically.Thus, J 1 \ J 2 =0.Toseethat J 1 J 2 =0,choose i 2 J i ( i =1 ; 2). Nowbytheconstructionof J 1 and J 2 ,wemaychooseaninteger m> 0,andelements ˘ 2 H 2 ( X 1 )and 2 H 2 ( X 2 ;M 2 )suchthat 1 = i ( ˘; ˘ )), k ( 2 )=( ( ) ; ). Then 1 2 = 1 m i ( ˘; ˘ )) 2 = 1 m [ ˘ ( ( )) ; ˘ ) ]=0.As 1 ; 2 werearbitrary, J 1 J 2 =0. Step2:Wereplace W byanother5-manifold W 1 with @W 1 = X , W 1 w W n i =1 S 2 and J = ker ( H 2 ( X ) ! H 2 ( W 1 )).Nowconsiderthecommutativediagram: 30 J ? ? y H 2 ( X 1 ) L H 2 ( X 2 ) i ! H 2 ( X ) @ ! H 1 ( M 1 ) ! 0 j ? ? y H 2 ( W 1 ) ? ? y 0 Byassumption, @ j J issurjectiveandquickdiagramchaseshowsthat = j i isalso. Thusif ˘ 2 H 2 ( W 1 ),thereareelements ˘ i 2 H 2 ( X i )( i =1 ; 2)suchthat ˘ = ( ˘ 1 ;˘ 2 ).But as i ( G ( J = Ker ( j ), ( ˘ 1 ; ˘ 1 ))= 1 ( ˘ 2 ) ; ˘ 2 )=0.Thus, ˘ = ( ˘ 1 ;˘ 2 )= 8 > > > < > > > : ( ˘ 1 ;˘ 2 )+ ( ˘ 1 ; ˘ 1 )) ( ˘ 1 ;˘ 2 )+ 1 ( ˘ 2 ) ; ˘ 2 ) = 8 > > > < > > > : (0 ;˘ 2 + ˘ 1 )) ( ˘ 1 + 1 ( ˘ 2 ) ; 0) : Clearlythisimpliesthat ˘ 2 image( H 2 ( X i ) ! H 2 ( W 1 ))( i =1 ; 2),andsobothho- momorphism H 2 ( X i ) ! H 2 ( W 1 )aresurjective.Sincerankof H 2 ( X i )=rankof H 2 ( W 1 ), i = j H 2 ( X i )( i =1 ; 2)areisomorphisms.Therefore, W 1 isarelativeh-cobordism.As G ( ker ( ), ' j H 2 ( X 1 ) = 1 2 1 ispreciselyByQuinn'srelativeh-cobordismthe- orem[36],thereexistsahomeomorphism F : X 1 ! X 2 suchthat F j M 1 = f and F = 3.3Plugs If X 1 and X 2 areclosedsimply-connectedsmooth4-manifolds,and Q X 1 ˘ = Q X 2 ,thenWall's theoremimplies X 1 and X 2 areh-cobordant,sothecorktheoremapplies.Nowletusconsider simply-connectedsmooth4manifoldswithboundaries.FromBoyer'sproof,weknowthatif X i areoddand( Q X 1 ;M 1 )isisomorphicto( Q X 2 ;M 2 ),then X 1 isrelativeh-cobordantto X 2 ,sotheybycorktwists.Whathappenswhen X i areeven?Theexampleinthelast 31 chapterwillshowthat X 1 maynotberelativeh-cobordantto X 2 .Inthissection,wewill provethefailureof X 1 beingrelativeh-cobordantto X 2 canbelocalizedona Y 1 of X 1 whichishomotopicequivalentto S 2 .Suchasubmanifoldiscalleda Plug .Plugs naturallyappearinmanyexoticmanifolds[9],[7],[42];itwasrstintroducedandstudied in[9]byAkbulutandYasui. Plugwasoriginallyin[9]asaStein4-manifold Y withboundary,homotopic equivalentto S 2 and ˝ : @Y ! @Y aninvolutionontheboundarysuchthat ˝ cannot extendtoanyself-homeomorphismof Y .Theprocedureofcutting Y andgluingitback by ˝ iscalled PlugTwist .Plugisasimilarobjectascork.Themainbetween themisplugtwistmaychangethehomeomorphismtype,whilecorktwistneverchanges thehomeomorphismtype.Wewanttoproveatheoremforplugsanalogoustothecork theorem.Tomakeourtheoremwork,weneedtoworkwithaweakerversionofplug.See theparagraphbelowthestatementofthetheorem. Theorem3.5. ( X 1 ;M 1 ) , ( X 2 ;M 2 ) aresimply-connectedsmooth4-manifoldswitho- morphicboundariessuchthat ( Q X 1 ;M 1 ) isisomorphicto ( Q X 2 ;M 2 ) .Then,thereexists submanifolds Y i ˆ X i ,( i =1 ; 2 )suchthat: (1) Y i arehomotopicequivalentto S 2 , @Y 1 ˇ @Y 2 ˇ ahomology S 1 S 2 . (2) X 1 n Y 1 ishomoemorphicto X 2 n Y 2 and i H 2 ( Y i ) ˆ i @ H 2 ( M i ) ,( i =1 ; 2 ),where i and i @ arethehomomorphismsinducedbytheinclusionmap i : Y i ! X i and i @ : M i ! X i . (3) Y i S id Y i = S 2 S 2 ,( i =1 ; 2 ); Y 1 S ˝ Y 2 = S 2 e S 2 ,where ˝ isanobviousomor- phismasweshallseeintheproof. (4) Y i canbemadeStein. Inourtheorem,weallow Y 1 and Y 2 beingtmanifolds,becauseaswehaveseenin 32 thecorktheorem, C 1 isnotnecessarilyto C 2 .Wecanmakethem byusing C 1 [ id C 1 ˇ S 4 ; C 1 [ ˝ C 2 ˇ S 4 .However,thistrickdoesnotworkforplugs,since plugsarenotcontractible. Proof. Glue X 1 , X 2 alongtheirboundariesby f . X 1 S f X 2 = X .ByNovikovAdditivity, ˙ ( X )=0;therefore, X boundsa5dimensionalmanifold W .Do1-surgerytokillall nontrivialelementsof ˇ 1 ( W ).Thus,wemayassume W issimplyconnected. Then,wecanusethe\handletrading"tricktocancel1-handlesand4-handles.Inthe end,wegetacobordismbetween X 1 and X 2 ,whichhasonly2-and3-handlesandinduces thetrivialcobordismbetween M 1 and M 2 (Itcanbeconsideredasthecollarof M 2 in X 2 ). Westillcallthiscobordism W .Considerthemiddlelevelwhichisbetweenthe2-handles andthe3-handles,callit X 0 , X 0 ' 1 ˇ X 1 ]nS 2 S 2 ]mS 2 e S 2 ' 2 ˇ X 2 ]nS 2 S 2 ]mS 2 e S 2 .We cannotproceedtheprooflikeCorkTheorem,becausewecannoteliminatetheexistenceof S 2 e S 2 .However,wecanassume m =1,since S 2 S 2 ] 2 S 2 e S 2 ˇ 2 S 2 S 2 ]S 2 e S 2 . Figure3.3:Handledecompositionofthecobordism In X 1 ]nS 2 S 2 ]S 2 e S 2 ,welet i ( i =1 ;:::;n )denotethebeltspheresof0framed5- dimensional2-handles; i denotethegeometricdualspheresof i ,i.e.,theattachingspheres generatedbythecoresofthe0framed2-handlesandthedisksboundedbytheirattaching 33 circles.[ i ] [ i ]=0,[ i ] [ i ]=0in H 2 ( X 1 ]nS 2 S 2 ]S 2 e S 2 ); i , j aredisjointwhen i 6 = j ; i intersectswith i geometricallyonce( i =1 ; 2 ;:::n ).Welet 0 denotethebelt sphereofthe1framed5-dimensional2-handle; 0 denotethegeometricdualsphereof 0 , i.e.,thespheregeneratedbythecoreofthe1framing2-handleandthediskboundedbyits attachingcircle.[ 0 ] [ 0 ]=0,[ 0 ] [ 0 ]=1in H 2 ( X 1 ]nS 2 S 2 ]S 2 e S 2 ); 0 intersects with 0 geometricallyonce.Byturningthehandlebodyupsidedown,wedenote 0 i , 0 i , 0 0 , 0 0 inasimilarmanner.Denoteby ' : X 1 ]nS 2 S 2 ]S 2 e S 2 ! X 2 ]nS 2 S 2 ]S 2 e S 2 the compositionof ' 1 ' 1 2 ,so ' 1 ( 0 i )aretheattachingspheresofthe5-dimensional3-handles i =0 ; 1 ;:::;n .Wewishtoprove ' ([ 0 ]) [ 0 0 ]=0; ' ([ 0 ]) [ 0 0 ]=1and ' ([ i ]) [ 0 i ]=1 for i =1 ;:::;n .Withoutambiguity,letusnotdistinguishbetween i , i andthehomology classes[ i ],[ i ]inthefollowingcontext. When X i areclosedmanifolds,byWall'sTheorem,wecana ˚ : X 1 ]nS 2 S 2 ]S 2 e S 2 ! X 1 ]nS 2 S 2 ]S 2 e S 2 suchthat ' ˚ = where agreeswithon H 2 ( X 1 ); i )= 0 i for i =1 ;:::;n and 0 )= 0 0 .Therefore,we canassumethat ' ( i )= 0 i , ' ( i ) 0 i =1, ' ( 0 ) 0 0 =0.When X i haveboundaries, Q X i arenotunimodular,sowecannotapplyWall'sTheoremdirectly.Weshallprove: ' ( 0 )= 0 0 + forsome 2 H 2 ( M 2 )and ' ( i )= i , i =1 ;:::n . If H 1 ( M 1 )isfree,considerthelongexactsequence: 0 ! H 2 ( M 1 ) i ! H 2 ( X 1 ]nS 2 S 2 ]S 2 e S 2 ) j ! H 2 ( X 1 ]nS 2 S 2 ]S 2 e S 2 ;M 1 ) ! H 1 ( M 1 ) ! 0. Assumetherankof H 2 ( X 1 ]nS 2 S 2 ]S 2 e S 2 )=m,therankof H 1 ( M 1 )=k.Let( Z m ;Q ) represent H 2 ( X 1 ]nS 2 S 2 ]S 2 e S 2 ),then( Z m ;Q )splitsas( Z m k ;Q 1 ) L ( Z k ; 0),where ( Z k ; 0)represents i H 2 ( M 1 );( Z m k ;Q 1 )represents coker ( j ).Easytocheck i ( a ) b =0 forany a 2 H 2 ( M 1 )and b 2 H 2 ( X 1 ]nS 2 S 2 ]S 2 e S 2 ),sothesplitisorthogonal,i.e., 34 Q = 0 B @ Q 1 0 00 1 C A ,and Q 1 isunimodular.Thesameistruefor H 2 ( X 2 ]nS 2 S 2 ]S 2 e S 2 ). ' inducesanisometry(anisomorphismwhichpreservestheintersectionform) ' : H 2 ( X 1 ]nS 2 S 2 ]S 2 e S 2 ) ! H 2 ( X 2 ]nS 2 S 2 ]S 2 e S 2 ).Notethat ' j i H 2 ( M 1 ) isan isometryon( Z k ; 0).Nowwewanttoanisometry~ ' :( Z m k ;Q 1 ) ! ( Z m k ;Q 1 ). Suppose ' ( i )= P m k j =1 p j j + P k j =1 q j j ,where f i ;i =1 ;:::;m k g isabasisof ( Z m k ;Q 1 ); f i ;i =1 ;:::;k g isabasisof( Z k ; 0).Let~ ' ( i )= P m k j =1 p j j .Clearly, ~ ' ' 1 ( i )= i ,so~ ' issurjective,andthereforeitisanisomorphismon coker ( j ).More- over,~ ' ( i ) ~ ' ( j )= ' ( i ) ' ( j )= i j .Thus,~ ' isanisometryon( Z m k ;Q 1 ). Let denotetheisometryon H 2 ( X 1 ]nS 2 S 2 ]S 2 e S 2 )whichagreewithon H 2 ( X 1 ), and i )= 0 i , i )= 0 i , i =1 ; 2 ;:::m ; 0 )= 0 0 , 0 )= 0 0 . Since Q 1 isunimodular,byWall'sTheorem,weareabletoda ˚ : X 1 ]nS 2 S 2 ]S 2 e S 2 ! X 1 ]nS 2 S 2 ]S 2 e S 2 suchthat ˚ ( x )= x forany x 2 i H 2 ( M 1 ) and ~ ˚ =~ ' 1 on coker ( j )( ~ ˚ isedinasimilarwayas~ ' ).Thiscanbeproved byworkingontherelativehandlebodypictures,similartotheproofin[30](ChparterX), whichdealswithhandlebodypicturesofclosedmanifolds.Therefore,( ' ˚ )( 0 )= 0 0 + forsome 2 i H 2 ( M 1 ).Thus,( ' ˚ )( 0 ) 0 0 =0;( ' ˚ )( 0 ) 0 0 =1;and ( ' ˚ )( i ) 0 i =1, i =1 ;:::;m ,wegetwhatweneed. If H 1 ( M 1 )isnotfree,weapplyBoyer'sresult. Letusdenote X i ]S 2 e S 2 by f X i ( i =1 ; 2), f X 1 [ f f X 2 by e X ;let id denotethe isomorphismfrom H 2 ( f X 1 )to H 2 ( f X 2 )thatagreeswithon H 2 ( X 1 )and id ( 0 )= 0 0 , id ( 0 )= 0 0 .Wenowconsiderthecobordism f W between f X 1 and f X 2 . 35 Figure3.4:Finerhandledecompositionofthecobordism ByBoyer'stheorem,wecan 0 i 2 H 2 ( e X )suchthat @ 0 i = i ,where i isabasisof thefreepartof H 1 ( M 1 )( i =1 ; 2 ;:::;k )and k ( 0 i ) 2 G ( id ) ).If 0 i 0 i isevenforall i ,weget ' ( 0 )=( id )( 0 )= 0 0 .Wearedoneinthiscase. If 0 p 0 p isoddforsome p ,assume k ( 0 p )=( a;b ),where b 2 H 2 ( f X 2 ;M 2 ), a =( id ) )( b ) 2 H 2 ( f X 1 ;M 1 ).Both H 2 ( f X 1 ;M 1 )and H 2 ( f X 1 )arefree. H 2 ( f X 1 ;M 1 ) a HOM ! ˇ H 2 ( f X 1 ;M 1 ) a PD ! ˇ H 2 ( f X 1 ) PD ( a ) . @ 0 p = p ) @ 1 a = p ,so a 2 1 ( H 1 ( M 1 ))where isthecoboundarymap, PD ( a ) 2 i ( H 2 ( M 1 ))anditisaprimitiveelementbyLefschetzduality. Let f x 1 = 0 ;x 2 = 0 ;x 3 = PD ( a ) ;x 4 ;:::;x m n g beabasisof H 2 ( f X 1 ).Notethat PD ( a ) 6 = 0 or 0 ,becauseweassumed 0 p 0 p isodd.Weconsideranautomorphism ˚ : H 2 ( f X 1 ) ! H 2 ( f X 1 )whichisby: ˚ ( 0 )= 0 + PD ( a ); ˚ ( x i )= x i for i 6 =1. Easytocheckthat ˚ preservestheintersectionform. Let f x 1 ; x 2 ; x 3 = a; x 4 ;:::; x m n g bethedualbasisfor H 2 ( f X 1 ;M 1 ),i.e., x i x j = ij . Then 0 = j 1 ( 0 ); 0 = j 1 ( 0 ) j 1 ( 0 ),because j 1 ( 0 ) 0 =0 ;j 1 ( 0 ) 0 = 1 ; ( j 1 ( 0 ) j 1 ( 0 )) 0 =1 ; ( j 1 ( 0 ) j 1 ( 0 )) 0 =0. 36 Claim3.3. ( ˚ ) 1 ( x i )= x i for i 6 =3 ; ( ˚ ) 1 ( a )= a 0 = a j 1 ( 0 )+ j 1 ( 0 ) ,where ( ˚ ) 1 isdinthefollowingcommutativediagram: H 2 ( f X 1 ) PD ! H 2 ( f X 1 ;M 1 ) HOM ! H 2 ( f X 1 ;M 1 ) ? ? y ˚ ? ? y ? ? y ( ˚ ) 1 H 2 ( f X 1 ) PD ! H 2 ( f X 1 ;M 1 ) HOM ! H 2 ( f X 1 ;M 1 ) Proof. ( ˚ ) 1 ( x i ) ˚ ( x j )= ij ,for i 6 =1 ; 2 ; 3, ˚ ( x i )= x i = ) ( ˚ ) 1 ( x i )= x i ; when i =1, 0 ˚ ( 0 )= 0 ( 0 + PD ( a ))=1, 0 ˚ ( 0 )= 0 0 =0, 0 ˚ ( PD ( a ))= 0 PD ( a )=0= ) ( ˚ ) 1 ( 0 )= 0 ; when i =2, 0 ˚ ( 0 )= 0 ( 0 + PD ( a ))=0, 0 ˚ ( 0 )= 0 0 =1, 0 ˚ ( PD ( a ))= 0 PD ( a )=0= ) ( ˚ ) 1 ( 0 )= 0 ; when i =3,( a 0 ) ˚ ( 0 )=( a 0 ) ( 0 + PD ( a ))=0,( a 0 ) ˚ ( 0 )=( a 0 ) 0 =0, ( a 0 ) ˚ ( PD ( a ))=( a 0 ) PD ( a )=1= ) ( ˚ ) 1 ( a )= a 0 . ˚ (*),since j 1 ( PD ( a ))=0: PD ( a ) j 1 ! 0 0 j 1 ! j 1 ( 0 )= 0 ? ? y ˚ ? ? y ( ˚ ) 1 ? ? y ˚ ? ? y ( ˚ ) 1 PD ( a ) j 1 ! 0 0 + PD ( a ) j 1 ! j 1 ( 0 )= 0 Therefore, ˚ (*). Nowweconsider f u p 0 = u 0 p i 1 ( 0 )+ i 1 ( 0 ), e u i 0 = u 0 i for i 6 = p ,then @ e u i 0 = @u 0 i = v i ; k ( e u i 0 )= k ( u 0 i ) k ( i 1 ( 0 ))+ k ( i 1 ( 0 ))= a;b ) j 1 ( 0 ) ; 0 + j 1 ( 0 ) ; 0 = ( ˚ ) 1 ( a ) ;b ).So, k ( e u i 0 ) 2 G ( ˚ ) 1 id ) and f u p 0 f u p 0 = u 0 p u 0 p 1whichis even.Thus, id ) ˚ 1 canberealizedgeometrically,i.e., ' agreeswith id ) ˚ 1 on H 2 ( f X 1 ). ' ( 0 )=( id ) ˚ 1 ( 0 )= 0 0 + forsome 2 H 2 ( M 2 ),so ' ( 0 ) 0 0 =0 and ' ( 0 ) 0 0 =1.Since f W isarelativeh-cobordism, ' ( i )= 0 i for i =1 ;:::;n .This canbeachievedbyintroducingcancellingpairsof5dimensional2-handlesand3-handles, andslidinghandles(see[40]1.7).Therefore, ' ( i ) 0 i =1. 37 Nowwecanputourhandsonconstructingtheplug. Letusassume n =0, f X 2 isphicto f X 1 .Doing2-surgeryon f X 2 along 0 0 (change\0"on 0 0 to\ ")givesus X 2 ;2-surgeryalong ' ( 0 )onthesamemanifoldyields X 1 . Weconsidertheneighborhoodof ' ( 0 ), 0 0 , 0 0 in f X 2 = X 2 ]S 2 e S 2 . Figure3.5:Theneighborhoodof ' ( 0 ), 0 0 , 0 0 in X 2 ]S 2 e S 2 f X 2 n nb ( ' ( 0 ))issimply-connected,because ' ( 0 )isatransversesphereof ' ( 0 ); f X 2 n nb ( 0 0 ) [ nb ( 0 0 )isalsosimply-connected.ByM-Vsequence,thehomologyof f X 2 n ( nb ( ' ( 0 )) [ nb ( 0 0 ) [ nb ( 0 0 ))isisomorphicto Z andgeneratedby a .ByVankampen Theorem,thefundamentalgroupof f X 2 n ( nb ( ' ( 0 )) [ nb ( 0 0 ) [ nb ( 0 0 ))isgeneratedbythe conjugatesof a .Any 2 thefundamentalgroupof f X 2 n ( nb ( ' ( 0 )) [ nb ( 0 0 ) [ nb ( 0 0 )) canberepresentedbyanembeddedloop,denoteitbyWechooseapoint 2 ' ( 0 ) nearanyintersectionpointbetween ' ( 0 )and 0 0 or 0 0 ,andthenwechooseanarc ˝ connectingthispointwithanypointonNowwedothemovefollowing ˝ ,and andthencomingbackbyanarcwhichisparallelandcloseenoughto ˝ .The 38 ofthismoveonthefundamentalgroupiskilling[ 1 ;b 0 ].Assume b 0 = r 1 ar , then[ a; ( r ) 1 ar ]=1.Since isarbitrary,[ s 1 as;t 1 at ]=[ a; ( ts 1 ) 1 ats 1 ]=1for any s;t 2 ˇ 1 ( f X 2 n ( nb ( ' ( 0 )) [ nb ( 0 0 ) [ nb ( 0 0 ))).Hence,aftermoves(creatingmore intersectionsbetween ' ( 0 )and 0 0 or 0 0 ),wecankillallthecommutators.Thefundamental groupnowbecomesabelian,therefore,itiscyclicandgeneratedby a . Foreachpairofintersectionpointsbetween ' ( 0 )and 0 0 withoppositesign,thereexists aWhitney'scircle l .Sincethefundamentalgroupof f X 2 n ( nb ( ' ( 0 )) [ nb ( 0 0 ) [ nb ( 0 0 ))is generatedby a ,wecanalwayschange l byadding2 ˇ twistsaroundthemeridianof ' ( 0 ) sothat l representsatrivialelementinthisfundamentalgroup. Figure3.6:FindingaWhitney'sdisk Therefore l boundsaWhitney'sdisk W disjointfrom ' ( 0 ), 0 0 and 0 0 .ThisWhitney's diskmightbeimmersed,thenwecanpushtheinteriorintersectionpointsof W at 0 0 .If theframingof W iswrong,wecandoboundarytwiststotheframingproblembycreating moreintersectionsbetween W and 0 0 .DoingWhitney'strickalongalltheseWhitney'sdisks, wecancanceltheintersectionpointsofoppositesignsbetween ' ( 0 )and 0 0 ,but 0 0 becomes animmersedsphere. 39 Figure3.7:Theneighborhoodof ' ( 0 ), 0 0 , 0 0 in X 2 ]S 2 e S 2 afterWhitney'smove Weaddall1-handlesof f X 2 intothispictureandintroduce2-and3-handlecancelling pairssuchthateach2-handlecancelsa1-handlealgebraically.Wedropallthe3-handlesand calltheresultingmanifold Y 0 . Y 0 issimply-connected.Doing2-surgeryon ' ( 0 )to Y 0 gives usasubmanifoldof X 1 ,callit Y 1 ;Doing2-surgeryon 0 0 to Y 0 givesusasubmanifoldof X 2 , callit Y 2 .Thehandledecompositionof Y 1 and Y 2 byexchanging\dot"and\0".That doesnotchangetheboundary3-manifold.So, @Y 1 ˝ ! @Y 2 isanobvioushomeomorphism. As 0 0 isanimmersedtransversespherefor 0 0 ' ( 0 ) ,afterdoing2-surgeryon 0 0 ' ( 0 ) , Y 2 ( Y 1 )isstillsimply-connected.Since X 1 n Y 1 hasidenticalhandledecompositionas X 2 n Y 2 , X 1 n Y 1 isto X 2 n Y 2 . Wewanttoshow Y i ( i =1 ; 2)arehomotopicequivalentto S 2 .Allofthe1-handlesand 2-handlesof Y 1 ( Y 2 )arealgebraicallycancelledexceptfor 0 0 ( ' ( 0 )),so Y 1 ( Y 2 )hasthe 40 samehomologygroupas S 2 .Byusingthelongexactsequence,wecaneasilycheckthat @Y 1 ( @Y 2 )isahomology S 1 S 2 .Ifweerase0framed2-handle 0 0 ( ' ( 0 ))fromthepicture, wegetacorkwhichiscontractible.Nowconsiderthemap g 1 : Y 1 ! S 2 ( g 2 : Y 2 ! S 2 ) suchthat g 1 ( g 2 )mapsthecomplementof 0 0 ( ' ( 0 ))in Y 1 ( Y 2 )toabasepointof S 2 andsqueezes 0 0 ( ' ( 0 ))ontoitscore.Thismapinducesanisomorphismonthehomology groups,therefore,byWhitehead'sTheorem, Y 1 ( Y 2 )ishomotopicequivalentto S 2 . Since H 2 ( Y 2 )isgeneratedby ' ( 0 )and ' ( 0 )= 0 0 + forsome 2 H 2 ( M 2 ), i H 2 ( Y 2 ) ˆ i @ H 2 ( M 2 ).Similarly, i H 2 ( Y 1 ) ˆ i @ H 2 ( M 1 ). Since Y 1 ( Y 2 )doesnotcontainany3-handles,thehandledecompositionof Y 1 [ id Y 1 can beconstructedfromthehandledecompositionof Y 1 byattachingeach2-handlea0framed 2-handlealongitsmeridianand3-handles.Thenumberof3-handlesattachedisthesame asthenumberof1-handlesof Y 1 .Thenbyhandleslides,everythingiscancelledexceptfor aHopflink( 0 0 anditsmeridian),framing0oneachcomponent.Clearly,thisis S 2 S 2 . Similarly, Y 2 [ id Y 2 is S 2 S 2 .Thehandledecompositionof Y 1 [ ˝ Y 2 isalmostthesame as Y 1 [ id Y 1 ,exceptthatthe2-handlewhichisattachedalongthemeridianof 0 0 nowis attachedalongthemeridianof ' ( 0 ).Byhandleslides,everythingiscancelledexceptfor 0 0 anditsmeridian,whichisclearlya S 2 e S 2 . Bydoingthe\cutandpaste"proceduresasin[1],wecanmake Y i Stein. If n> 0,afterdoing2-surgeryon X 2 ]nS 2 S 2 ]S 2 e S 2 along 0 0 ,wewillget X 2 ]nS 2 S 2 ; doingsurgeryonthesamemanifoldalong ' ( 0 )willgiveus X 1 ]nS 2 S 2 .Ifwekeepdoing surgeryalong 0 i ( ' ( i )),( i =1 ;:::;n ),wewillobtain X 2 ( X 1 ).So, X 2 and X 1 by n +1\dot"and\0"exchanges.If ' ( 0 )and 0 i ( i =1 ;:::;n )areunlinked; 0 0 and ' ( i ) ( i =1 ;:::;n )areunlinked,thenexchanging\dot"and\0"between ' ( 0 )and 0 0 isa plugtwist;exchanging\dot"and\0"between i and 0 i ( i =1 ;:::;n )are n corktwists. 41 Otherwise,WLOG,wemayassume ' ( 0 )and 0 j ( j =1 ;:::;l )arelinked,thenwewill considertheneighbourhoodof ' ( 0 ) S [ l j =1 ' ( j ) S 0 0 S [ l j =1 0 j in X 2 ]nS 2 S 2 ]S 2 e S 2 . Weincludeallthe1-handlesandaddalgebraiccancelling2-handles.Wecalltheresulting manifold Y 0 .Doingsurgeryon Y 0 along ' ( 0 )and ' ( j )( j =1 ;:::;l )willgiveus Y 1 ;doing surgeryon Y 0 along 0 0 and 0 j ( j =1 ;:::;l )willgiveus Y 2 .( X 1 n Y 1 ) ] ( n l ) S 2 S 2 is to( X 2 n Y 2 ) ] ( n l ) S 2 S 2 sincetheirhandledecompositionareidentical. Therefore,( X 1 n Y 1 )and( X 2 n Y 2 )bycorktwistsandhencetheyarehomeomorphic. Itisalsoeasytocheckthat Y 1 and Y 2 satisfyotherdesiredproperties. Notethat i ( H 2 ( Y i )) ˆ H 2 ( M i )( i =1 ; 2)guaranteesthattheplugtwistdoesnotchange theintersectionformof X i ,andkeepholdof(*).Whileif i ( H 2 ( Y 2 ))=0in H 2 ( X 2 ), then ' ( 0 )=0in H 2 ( X 2 ).Ontheotherhand,byusingBoyer'stheorem,wecan 0 i 2 H 2 ( X )( X = X 1 [ f X 2 )suchthat @ 0 i = i and k ( 0 i ) 2 G ( ),where i isabasis ofthefreepartof H 1 ( M 1 ), i =1 ;:::;m n 1.If 0 i 0 i isevenfor 8 i =1 ;:::;m n 1, then X 1 ishomeomorphicto X 2 .Otherwise, 0 p 0 p isoddforsome p ,then i ( 0 p ) i ( 0 p ) isalsoodd,where i isthehomomorphisminducedbytheinclusionmap i : X ! e X = X]S 2 e S 2 ] S 2 e S 2 .Obviously, @ ( i ( 0 i ))= i and k ( i ( 0 i )) 2 G ( id ) ).Sobythe constructionintheproof, ' ( 0 )= 0 0 + forsomenon-trivialelement 2 H 2 ( M 2 ) ˆ H 2 ( X 2 ).Acontradiction.Hence: Corollary3.1. Undertheassumptionoflasttheorem, X 1 ishomeomorphicto X 2 if Y 2 is null-homologousin X 2 ,or Y 1 isnull-homologousin X 1 ,. Byproperty(3)ofthetheorem,wecouldthinktheplugtwist X 1 ! X 2 ascut- tingasubmanifold Y 1 whichisahomotopic S 2 D 2 from X 1 andgluingbackthe complementof Y 1 in S 2 e S 2 bythenaturallyinducedboundaryhomeomorphism @Y 1 ! 42 @ (complementof Y 1 )in S 2 e S 2 .Thiscanbeconsideredasageneralizationof\GluckTwist" whichisanoperationthatcut S 2 D 2 from X 1 andgluebackthecomplementof S 2 D 2 in S 2 e S 2 (whichisstilla S 2 D 2 )bythenaturallyinducedboundaryhomeomorphism S 2 S 1 ! S 2 S 1 . [9],[14]giveaveryeasydescriptionofthehandle-bodypictureofGlucktwist.Exchanging \dot"and\0"inthefollowingpictureisaGlucktwist. Figure3.8:Glugtwist PlugtwistcanbeviewedasageneralizationoftheGlucktwistalsofromitshandle-body picture.Recallthatintheproof,assuming n =0,beforewedosurgeryalong ' ( 0 )and 0 0 wehavethefollowingsubmanifoldof f X 2 43 Figure3.9:Theneighborhoodof ' ( 0 ), 0 0 , 0 0 in X 2 ]S 2 e S 2 Afterthemoves,thefundamentalgroupof f X 2 n ( nb ( ' ( 0 )) [ nb ( 0 0 ))becomes trivial.SoeachWhitney'scircleconnectingapairofintersectionpointsbetween ' ( 0 ) and 0 0 withoppositesignboundsanimmersedWhitney'sdiskdisjointfrom ' ( 0 )and 0 0 (itmightintersectwith 0 0 ).Thelinkingtorus T 2 intersectswiththisWhitney'sdisk algebraicallyonce,and[ T 2 ] [ T 2 ]=0,sobyCasson'stheorem,thereisaCassonhandle attachedalongthisWhitney'scircle.ThenbyFreedman'stheorem,alltheseCassonhandles arehomeomorphictostandard2-handlesrelativetheboundary.Therefore,thereexistsa manifold Z homeomorphicto f X 2 suchthatin Z ,theseCassonhandlesbecomestandard 2-handles.SincethishomeomorphismonlychangetheinterioroftheCassonhandle, ' ( 0 ) and 0 0 arepreservedbythishomeomorphism. 0 0 isalsopreservedbecauseitintersectswith theCassonhandlesonlyatpoints.So,in Z ,theintersectionpointsbetween ' ( 0 )and 0 0 withoppositesignscanbecancelled,while ' ( 0 )mayhavemoreintersectionpointswith 0 0 thanin f X 2 .Doingsurgeryalong ' ( 0 )on Z givesamanifold X 0 1 homeomorphicto X 1 . Similarly,doingsurgeryalong 0 0 on Z givesamanifold X 0 2 homeomorphicto X 2 . 44 Corollary3.2. Undertheassumptionofthelasttheorem,thereexists X 0 i suchthat X 0 i is homeomorphicto X i ( i =1 ; 2 )and X 0 1 ;X 0 2 arerelatedbydoingthefollowing\dot"and\0" exchange. Figure3.10:Plugtwist Thetypechangefrom X i to X 0 i isduetocorktwists;the\dot"and\0" exchangeintheabovepictureistheprimecauseofchangingthehomeomorphismtype.It iseasytoobservethat:if ' ( )isunlinkedwith 0 0 ,itisaGlucktwist;ifthereisoneclasp between ' ( )and 0 0 (asintheabovepicture),itisa0logarithmictransformation.Since weallowanynumberofclaspsbetween ' ( )and 0 0 .Thisoperationcanbethoughtofa generalizationofGlucktwistand0logarithmictransformation. 45 Figure3.11:Ageneralplugtwist Combinethecorktheoremstatedinthesectionofthischapter,wecansummarize thefeaturesofcorkandplugbythefollowingpictures: If X 1 and X 2 aresimply-connectedsmooth4manifoldssuchthat( Q X 1 ;@X 1 )isisomor- phicto( Q X 2 ;@X 2 ),then X 1 and X 2 arerelatedby: Oneplugtwist ! changeshomeomorphismtype Corktwists ! changetype Figure3.12:Corktwiststogetherwithaplugtwist 46 Chapter4 AnExample 4.1Anirreducible3manifold Foraclosed,connected,orientable3manifold M ,denote H ( M )thegroupofisotopyclasses ofhomeomorphismsof M ; H + ( M )thethegroupofisotopyclassesoforientationpreserving homeomorphismsof M . RecallBoyer'sresult:( X 1 ;@X 1 )and( X 2 ;@X 2 )( @X 1 ˇ @X 2 ˇ M )aresimply-connected topological4-manifolds,if( Q X 1 ;@X 1 )isisomorphicto( Q X 2 ;@X 1 )through ;f ),thenthey arehomeomorphicifoneofthefollowingconditionsholds: (i) H 1 ( M; Q )=0and X 1 ) X 2 ); (ii) X i areoddmanifoldsand X 1 ) X 2 ); (iii) X i areevenmanifoldsand X 1 [ g X 2 isalsoevenforsomehomeomorphism g : @X 1 ! @X 2 . Notethatinthelastcase, X 1 [ f X 2 mightbeodd,butifwecouldaself-homeo- morphism h : @X 2 ! @X 2 suchthat h interchangesthetwospinstructureson @X 2 induced from f ( @X 1 )and X 2 respectively,then X 1 [ h f X 2 iseven,andthus X 1 ishomeomorphic to X 2 .Conversely,ifthereexistsahomeomorphism F : X 1 ! X 2 ,then F j @X 1 f 1 must interchangesthetwospinstructureson @X 1 .Sothissuggestsus,foraboundary M andaintersectionform Q presenting M ,thereareatmostSpin( M ) = H + ( M )t homoemorphismtypesofsimply-connectedtopological4-manifoldswithboundary M and 47 intersectionform Q .Wedenotethesetofthesemanifoldsby V Q ( M ). Theorem4.1. (Boyer,[15]) When H 1 ( M ) isfree, V Q ( M ) isone-to-onecorrespondentto: (i)Spin ( M ) = H + ( M ) if Q iseven; (ii) Z = 2 if Q isodd. Forinstance,when M = S 1 S 2 ,fora Q presenting S 1 S 2 , jV Q ( S 1 S 2 ) j = 8 > > > < > > > : 1 ; if Q iseven 2 ; if Q isodd(thetwohomeomorphismtypesby : ,since H + ( S 1 S 2 )actstransitivelyonSpin( S 1 S 2 ).([25]) When M = T 3 ,weknowthat T 3 has8tspinstructures; H + ( T 3 )isisomorphic to SL (3; Z )whichisgeneratedby6elements. H + ( T 3 )actstransitivelyon7spinstructures, howeverthereexistsanexceptionalspinstructurewhichisonatorbitbyitself, denotedby s Lie ([30],ChapterIV).Thisspinstructurespin-boundsthecomplementofa genericerintherationalelliptic E (1)= CP 2 ] 9 CP 2 .([30],ChapterV) Therefore, jV Q ( T 3 ) j = 8 > > > < > > > : 2 ; if Q iseven 2 ; if Q isodd(thetwohomeomorphismtypesby : So,thereexist2simply-connectedeventopological4-manifolds X 1 , X 2 suchthat @X 1 = @X 2 = T 3 ,( Q X 1 ;T 3 )isisomorphicto( Q X 2 ;T 3 )and X 1 isnothomeomorphicto X 2 . WLOG,weassumetheinducedspinstructureon @X 1 is s Lie ,whiletheinducedspinstruc- tureon @X 2 isoneoftheother7spinstructures. SinceBoyer'sconstructioninvolvesFreedman'stheorem: Everyhomology3-sphereboundsacontractibletopological4-manifolds . WeknowthatFreedman'stheoremfailsforsmooth4-manifolds,sothesecondpartofBoyer's resultmaynotbetrueinthesmoothsituation. 48 Indeed,theRohlininvariantof( T 3 ;s Lie )is8,since ˙ ( CP 2 ] 9 CP 2 )= 8, ˙ ( T 2 D 2 )=0, ˙ (thecomplementof T 2 D 2 )= 8byNovikovadditivity;theRohlininvariantofthe other7spinstructureson T 3 is0,sincetheyallbound T 2 D 2 .Sotheredonotexist smooth4-manifolds X 1 , X 2 suchthat X 1 induces s Lie on @X 1 ˇ T 3 , X 2 inducesoneofthe other7spinstructureson @X 2 and Q X 1 ˘ = Q X 2 ˇ T 3 .Thus,if X 1 and X 2 aresmooth simply-connected4-manifoldssuchthat( Q X 1 ;T 3 )isisomorphicto( Q X 2 ;T 3 ),then X 1 is homeomorphicto X 2 . Itisnaturaltoaskthequestion:Isitalwaystruethatfortwosimply-connectedsmooth 4-manifolds X 1 and X 2 ,( Q X 1 ;M )isomorphicto( Q X 2 ;M )impliestheyarehomeomorphic? Wewillshowthatthefollowingexamplegivesanegativeanswer: ˇ Figure4.1:Asimply-connected4-manifold 49 ˇ Figure4.2:DoingGlucktwistgivesanon-homeomorphicsimply-connected4-manifold Clearly, Q X 1 ˘ = Q X 2 , @X 1 ˇ @X 2 ;denotetheboundaryby M .Wecaneasilythe plugtwist;itisindeedaGlucktwist. H 2 ( Y i ) ˆ H 2 ( @X i )( i =1 ; 2),soeasytocheckthat (*)iscommutative,i.e.,wehave( Q X 1 ;M )isomorphicto( Q X 2 ;M ).Toprove X 1 isnot isomorphicto X 2 ,weonlyneedtoprovetheredoesnotexistahomeomorphism M ! M interchangingthetwospinstructuresinducedfrom X 1 and X 2 respectively. 50 Figure4.3:Findingthefundmentalgroupoftheboundary WecaneasilywritetheWirtingerpresentationofthefundamentalgroupof M .Ithas 20generators: f x 1 ;:::;x 4 ;y 1 ;:::;y 10 ;a 1 ;b 1 ;a 2 ;b 2 ;a 3 ;b 3 g and25relations: y 8 = a 3 y 8 b 1 3 ; y 9 = a 3 y 9 b 1 3 ; y 7 = a 3 y 8 a 1 3 ; y 10 = a 3 y 9 a 1 3 ; y 6 = a 2 y 6 b 1 2 ; y 10 = a 2 y 10 b 1 2 ; y 5 = a 2 y 6 a 1 2 ; y 4 = a 2 y 10 a 1 2 ; y 2 = a 1 y 2 b 1 1 ; y 4 = a 1 y 4 b 1 1 ; y 1 = a 1 y 2 a 1 1 ; y 3 = a 1 y 4 a 1 1 ; x 1 = y 1 x 1 y 1 3 ; x 4 = y 1 x 1 y 1 1 ; x 4 = y 2 x 3 y 1 2 ; x 3 = y 2 x 3 y 1 5 ; x 3 = y 7 x 3 y 1 6 ; x 2 = y 7 x 3 y 1 7 ; x 2 = y 8 x 1 y 1 8 ; x 1 = y 8 x 1 y 1 9 ; y 9 y 1 8 =1; y 10 y 1 6 =1; y 4 y 1 2 =1; y 1 8 y 7 y 1 2 y 1 =1; x 1 1 a 1 3 x 1 a 1 2 x 1 1 a 1 1 x 1 a 1 a 2 a 3 =1. The3equationsofthelastlineimplies y 8 = y 9 , y 2 = y 4 , y 6 = y 10 ,substitutinginto thelastequationsofthe3rows,weget y 6 = a 3 y 8 a 1 3 , y 2 = a 2 y 6 a 1 2 , y 3 = a 1 y 2 a 1 1 .By comparingthesecondlastequationsofthest3rows,wehave y 7 = y 6 ;y 2 = y 5 ;y 1 = y 3 . 51 So, y 1 = y 3 ;y 2 = y 4 = y 5 ;y 6 = y 7 = y 10 ;y 8 = y 9 ; x 1 = x 2 = x 3 = x 4 .Hence,after weget: ˇ 1 ( M )= h x 1 ;y 1 ;a 1 ;a 2 ;a 3 j [ x 1 ;y 1 ]=1 ; [ x 1 ;y 2 ]=1 ; [ x 1 ;y 6 ]=1 ; [ x 1 ;y 8 ]=1 ; y 1 8 y 6 y 1 2 y 1 =1( y ) ;x 1 7 x 6 x 1 5 x 1 =1( z ) i , where y 2 = a 1 1 y 1 a 1 ; y 6 = a 1 2 a 1 1 y 1 a 1 a 2 ; y 8 = a 1 3 a 1 2 a 1 1 y 1 a 1 a 2 a 3 ; x 5 = a 1 x 1 a 1 1 ; x 6 = a 1 a 2 x 1 a 1 2 a 1 1 ; x 7 = a 1 a 2 a 3 x 1 a 1 3 a 1 2 a 1 1 . Nowwedenote ˇ 1 ( M )by G ,Let G 1 = h x;a;b;c j cx 1 c 1 bxb 1 ax 1 a 1 x =1 i ; G 2 = h y;a;b;c j c 1 y 1 cb 1 yba 1 y 1 ay =1 i , where x = x 1 ;y = y 1 ;a = a 1 ;b = a 1 a 2 ;c = a 1 a 2 a 3 .Easytoseethat G = h G 1 ;G 2 i ,the groupgeneratedby G 1 and G 2 ,and G 1 \ G 2 6 = f 1 g . Wecancheckthattherelatorof G 1 , r 1 = cx 1 c 1 bxb 1 ax 1 a 1 x isofminimallength under Aut ( h a;b;c;x i )byWhitehead'sTheorem;similarly,therelatorof G 2 , r 2 = c 1 y 1 c b 1 yba 1 y 1 ay isofminimallengthunder Aut ( h a;b;c;y i ). Theorem4.2. (Whitehead,[46]) If w , v 2 F n suchthat w canbetransformedto v by automorphismsof F n ,and v isofminimallength,thenthereexistsasequence S 1 ;S 2 ;:::;S m ofTypeIandTypeIIautomorphismssuchthat S m :::S 2 S 1 ( w )= v ,andfor k m , j S k :::S 2 S 1 ( w ) j S k 1 :::S 2 S 1 ( w ) j ,withstrictlyinequalityunless S k 1 :::S 2 S 1 ( w ) ismin- imal. Ifwedenotethesetofgeneratorsandtheirreverses f x 1 ;x 2 ;:::x n ;x 1 1 ;:::;x 1 n g of F n by L n ,thenWhiteheadTypeIautomorphismisapermutation S 2 Aut F n actingon L n such that S ( x 1 )=( S ( x )) 1 for 8 x 2 L n ; 52 For x 2 L n and A 2 L n ,WhiteheadTypeIIautomorphism S ( A;x )isasfollows: S ( y )= 8 > > > > > > > > > > > > < > > > > > > > > > > > > : yx; if y 2 A;y 1 62 A;y 62f x;x 1 g ; x 1 y; if y 62 A;y 1 2 A;y 62f x;x 1 g ; x 1 yx; if y;y 1 2 A; y; otherwise. forany y 2 L n . Now,byusingthefollowingtheorem, Theorem4.3. ([31],Proposition5.13) Let H = h x 1 ;x 2 ;:::;x n j r i where r isofminimal lengthunder Aut ( h x 1 ;x 2 ;:::;x n i ) andcontainsexactlythegenerators x 1 ;x 2 ;:::;x k forsome 0 k n .Then H = H 1 H 2 where H 1 = h x 1 ;x 2 ;:::;x k j r i isfreelyindecomposableand H 2 isfreewithbasis x k +1 ;:::;x n . Agroup G iscalled freelyindecomposable if G cannotbewrittenas G = A B for nontrivialsubgroups A G;B G . wecaneasilyseethat: Lemma4.1. G 1 and G 2 arefreelyindecomposable. Proof. Bythistheorem, G 1 cansplitsas G 1 = A 1 B 1 where A 1 isfreelyindecomposable and B 1 isfree.Since r 1 = cx 1 c 1 bxb 1 ax 1 a 1 x , k = n =4.Therefore, B 1 istrivial, and G 1 = A 1 whichisfreelyindecomposable.Similarly,onecanprovethat G 2 isalsofreely indecomposable. Nextwewillshowthat G isalsofreelyindecomposablebyapplyingKuroshsubgroup theorem. Theorem4.4. (KuroshSubgroupTheorem) :If G isfreelydecomposable,i.e., G = A B where A;B arebothnontrivialsubgroups,and H G isasubgroupof G ,thenthereexista 53 family ( A i ) i 2 I ofsubgroups A i A ,afamily ( B j ) j 2 J ofsubgroups B j B ,families g i ;i 2 I and f j ;j 2 J ofelementsof G ,andasubset X G suchthat H = F ( X ) ( ( i 2 I ) g i A i g 1 i ) ( ( j 2 J ) f j B j f 1 j ) ,where F ( X ) isthefreegroupgeneratedby X . Proposition4.1. G isfreelyindecomposable. Proof. Since G 1 isnotfreeandisfreelyindecomposable,byKuroshsubgrouptheorem, G 1 = gA 1 g 1 or G 1 = gB 1 g 1 forsome g 2 G and A 1 A , B 1 B ;similarly, G 2 = fA 2 f 1 or G 2 = fB 2 f 1 forsome f 2 G and A 2 A , B 2 B .If G 1 = gA 1 g 1 and G 2 = fB 2 f 1 , then a = gsg 1 = frf 1 forsome s 2 A 1 A and r 2 B 2 B ,as a 2 G 1 \ G 2 .This contradictsto G = A B .Similarly, G 1 = gB 1 g 1 , G 2 = fA 2 f 1 isnotpossible. Nowsuppose G 1 = gA 1 g 1 and G 2 = fA 2 f 1 .Ifboth f and g 2 A ,then G 1 A ; G 2 A ,so G A , B istrivial.Ifoneof f and g isnotcontainedin A ,say f involves non-trivialelementsof B ,but g doesnot,then a 2 G 1 A and a 2 G 2 sothat a = frf 1 , contradictingwith G = A B .Nowifneither f nor g iscontainedin A ,wetakeanynon- trivialelement s 2 A , s 2 G = h G 1 ;G 2 i ,so s isequaltoaproductof ft i f 1 and gt j g 1 forsome t i ;t j 2 A .As f and g involvenon-trivialelementsof B ,thiscontradictswith G = A B .Thecasewhere G 1 = gB 1 g 1 and G 2 = fB 2 f 1 canbeprovedinasimilar manner. Thispropositionimpliesthat M isaprimemanifold,becauseif M = M 1 ]M 2 ,then ˇ 1 ( M )= ˇ 1 ( M 1 ) ˇ 1 ( M 2 )byVankampenTheorem.Clearly, M isnota2-spherebundle over S 1 ,soitisanirreducible3-manifold.Astherankof H 1 ( M )is5, M istly large. 54 4.2Finding H + ( M ) by Out ( ˇ 1 ( M )) Denote Out ( ˇ 1 ( M ))theouterautomorphismgroupofthefundamentalgroupof M .Wald- hausen[43]provedthat: When M isclosedorientableirreducible3-manifoldwhichislarge,thenaturally inducedhomomorphism : H ( M ) ! Out ( ˇ 1 ( M )) isanisomorphism. Wewillanalyse Out ( ˇ 1 ( M ))andthecorrespondingorientation-preservinghomeomor- phisms.Wewilleasilyseethatnoneofthesehomeomorphismscaninterchangethetwospin structures.Therefore X 1 isnothomeomorphicto X 2 . Lemma4.2. If isanautomorphismof G ,thenatleastoneof x 1 ) and y 1 ) iscontained in h Cl ( x 1 ) ;Cl ( y 1 ) i ,where Cl ( x 1 )= f g 1 x 1 g : g 2 G g ; Cl ( y 1 )= f g 1 y 1 g : g 2 G g arethe conjugateclassesof x 1 and y 1 . Proof. Denote H thenormalsubgroupgeneratedby g 1 x 1 1 gx 1 and g 1 y 1 1 gy 1 for 8 g 2 G , then G=H = h x 1 ih y 1 ih a 1 ;a 2 ;a 3 i .Let x denotetheimageof x inthequotient G=H .In G=H , x 1 ; y 1 commutewithallelements,so x 1 )= x m 1 1 y n 1 1 f 1 ; y 1 )= x m 2 1 y n 2 1 f 2 forsome f 1 ;f 2 2h a 1 ;a 2 ;a 3 i .Assumethat x 1 ) = 2h Cl ( x 1 ) ;Cl ( y 1 ) i ; y 1 ) = 2h Cl ( x 1 ) ;Cl ( y 1 ) i ,then f 1 and f 2 arenottrivialelements. Now, x 1 ) y 1 )= x m 1 + m 2 1 y n 1 + n 2 1 f 1 f 2 ; x 1 ) y 1 )= x m 1 + m 2 1 y n 1 + n 2 1 f 2 f 1 .In G , x 1 y 1 = y 1 x 1 ,so x 1 y 1 )= y 1 x 1 ).Hence f 1 f 2 = f 2 f 1 .Considerthesubgroup K of h a 1 ;a 2 ;a 3 i generatedby f 1 ;f 2 .Anysubgroupofafreegroupisfree,so K isafreegroup. Byconsideringtheabelinizationof K ,itiseasytoseethattherankof K 6 2,since K isgeneratedby2elements.Iftherankof K is2,asaconsequenceofGrushko'sTheorem, f 1 ;f 2 generate K freely,whichcontradictstothefact f 1 f 2 = f 2 f 1 .Thus,therankof K =1, whichimplies f 1 = t k 1 ;f 2 = t k 2 forsome t 2h a 1 ;a 2 ;a 3 i and k 1 , k 2 2 Z . 55 Assume a 1 )= x m 3 1 y n 3 1 f 3 forsome f 3 2h a 1 ;a 2 ;a 3 i ,then x 1 a 1 ) 1 y 1 a 1 )= x m 1 + m 2 + m 3 1 y n 1 + n 2 + n 3 1 f 1 f 1 3 f 2 f 3 ; while a 1 ) 1 y 1 a 1 x 1 )= x m 1 + m 2 + m 3 1 y n 1 + n 2 + n 3 1 f 1 3 f 2 f 3 f 1 . x 1 a 1 1 y 1 a 1 = a 1 1 y 1 a 1 x 1 in G ,so x 1 a 1 ) 1 y 1 a 1 )= a 1 ) 1 y 1 a 1 x 1 ), f 1 f 1 3 f 2 f 3 = f 1 3 f 2 f 3 f 1 .Thereforetherankof h f 1 ;f 1 3 f 2 f 3 i =1= ) rankof h f 1 ;f 2 ;f 3 i = 1.Thus,wecanconclude f 1 = t k 1 ;f 2 = t k 2 ;f 3 = t k 3 forsome t 2h a 1 ;a 2 ;a 3 i andintegers k 1 ;k 2 ;k 3 . x 1 a 1 2 a 1 1 y 1 a 1 a 2 = a 1 2 a 1 1 y 1 a 1 a 2 x 1 ; x 1 a 1 3 a 1 2 a 1 1 y 1 a 1 a 2 a 3 = a 1 3 a 1 2 a 1 1 y 1 a 1 a 2 a 3 x 1 in G ,sowecansimilarlyprovethat a 2 )= x m 4 1 y n 4 1 t k 4 ; a 3 )= x m 5 1 y n 5 1 t k 5 . Nowweconsider G= [ G;G ]= h x 1 ih y 1 ih a 1 ih a 2 ih a 3 i ,whichisafreeabeliangroup ofrank5.Theabelinizationof G isthesameastheabelinizationof G=H . G isgenerated by f x 1 ) ; y 1 ) ; a 1 ) ; a 2 ) ; a 3 ) g ,butrankof h x 1 ) ; y 1 ) ; a 1 ) ; a 2 ) ; a 3 ) i [ G;G ] 6 3 : Acontradiction.Therefore,atleastoneof x 1 )and y 1 ) 2h Cl ( x 1 ) ;Cl ( y 1 ) i . Lemma4.3. If I := f sxs 1 ;s 2h a 1 ;:::;a n ig isafreegroup, u;v 2 I , v isnontrivial,and uv k = dv l d 1 u forsome d 2h a 1 ;:::;a n i andintegers k and l ,then d =1 and uv = vu . Proof. If u istrivial,Itisobvious.Letusassume u isnottrivial. If v isnotcyclicallyreduced, v = sv 0 s 1 forsomenontrivialelement s 2 I ,where v 0 iscyclicallyreduced,i.e.,theletterof v 0 isnottheinverseofthelastletterof v 0 .Then usv k 0 s 1 u 1 = dsd 1 dv l 0 d 1 ds 1 d 1 .Thus, ds 1 d 1 usv k 0 s 1 u 1 d 1 dsd 1 = dv l 0 d 1 .Denote ds 1 d 1 us by u 0 ,wehave u 0 v k 0 u 1 0 = dv l 0 d 1 .Therefore,WLOG,wecan assumethat v iscyclicallyreduced. 56 Since uv k u 1 = dv l d 1 , u or u 1 mustbecompletelycancelledouttomake dv l d 1 cyclicallyreduced( d 2 I ).Ifthelastletterof u cancelstheletterof v ,thenthe letterof u 1 cannotbecancelledwiththelastletterof v andviceverse,since v iscyclicallyreduced.Therefore, k = l .Assume v k = e 1 xe 1 1 e 2 xe 1 2 :::e m xe 1 m ,where e i 2h a 1 ;:::;a n i ,then u = e p x 1 e 1 p e p 1 x 1 e 1 p 1 :::e 1 x 1 e 1 1 forsome pl ( t 2 ),then t 1 3 = t 1 2 ct 2 t 1 4 forsome t 4 2h a;b;c i ,so t 1 4 cbat 4 t 1 2 c 1 t 2 a 1 = t 1 1 bt 1 .Notethatthelastletterof t 2 is a ,wedenote t 2 a 1 by t 5 .Thenwehave t 1 4 cbat 4 a 1 t 1 5 c 1 t 5 = t 1 1 bt 1 ,andthereisnocancellationbetween t 1 4 and a 1 .Clearly, l ( t 4 )+1 6 = l ( t 5 ),sinceotherwisethemiddleletteronthelefthandsideis a 1 ,whilethe middleletterontherighthandsideis b . (1aa)Now,if l ( t 4 )+1 >l ( t 5 ), t 1 = t 6 t 1 5 c 1 t 5 forsome t 6 2h a;b;c i ,thenwehave t 1 4 cbat 4 = t 1 5 ct 5 t 1 6 bt 6 a .Sowearebacktotheoriginalfrom: t 4 = t 3 t 1 2 ct 2 playstherole of t 3 ; t 5 = t 2 a 1 playstheroleof t 2 ; t 6 = t 1 at 1 2 ct 2 a 1 playstheroleof t 1 andweknow that l ( t 4 ) l ( t 1 ),thenbyconjugatingbothsideswith a , at 1 3 cbat 3 a 1 = at 1 2 ct 2 t 1 1 bt 1 .Nowweknow t 3 a 1 = t 4 t 1 1 bt 1 forsome t 4 2h a;b;c i .Therefore, a 1 t 1 1 b 1 t 1 t 1 4 cbat 4 = t 1 2 ct 2 : (1ba)If l ( t 1 ) l ( t 4 )+1, t 2 = t 5 t 1 4 cbat 4 forsome t 5 2h a;b;c i .Thenwehave a 1 t 1 1 b 1 t 1 = t 1 4 ( cba ) 1 t 4 t 1 5 ct 5 ,whichimplies t 1 4 ( cba ) t 4 = t 1 5 ct 5 t 1 1 bt 1 a .So t 4 = t 3 a 1 t 1 1 b 1 t 1 playstheroleof t 3 ; t 5 = t 2 t 1 4 ( cba ) 1 t 4 playstheroleof t 2 ; t 1 doesnot changeand l ( t 4 ) l ( t 2 ),then t 1 1 = t 1 2 c 1 t 2 t 1 5 forsome t 5 2h a;b;c i ,so at 1 5 bt 5 t 1 2 ct 2 = t 1 4 cbat 4 .Note,asweassumethelastletterof t 1 isneither a or a 1 ,eitheris t 2 by( ?? ).So thestletterof t 1 5 mustbe a 1 .Theequationcanwrittenas t 1 6 bt 6 at 1 2 ct 2 = t 1 4 cbat 4 , where t 6 = t 5 a 1 .Clearly, l ( t 2 ) 6 = l ( t 6 ). (2aa)If l ( t 6 ) >l ( t 2 ), t 4 = t 7 t 1 2 ct 2 forsome t 7 2h a;b;c i .Thenwehave t 1 6 bt 6 a = t 1 2 c 1 t 2 t 1 7 cbat 7 = ) t 1 7 cbat 7 = t 1 2 ct 2 t 1 6 bt 6 a .Now t 7 = t 3 a 1 t 1 2 c 1 t 2 playstheroleof t 3 ; t 2 doesnotchange; t 6 = t 1 t 1 2 c 1 t 2 a 1 playstheroleof t 1 and l ( t 7 ) l ( t 6 ), t 2 = t 7 t 1 6 bt 6 forsome t 7 2h a;b;c i ,so t 1 4 ( cba ) 1 t 4 a = t 1 6 b 1 t 6 t 1 7 c 1 t 7 = ) t 1 3 cbat 3 = t 1 7 ct 7 t 1 6 bt 6 a .Now, t 3 doesnotchange; t 7 = t 2 t 1 6 b 1 t 6 playstheroleof t 2 ; t 6 = t 1 t 1 4 ( cba ) 1 t 4 a playstheroleof t 1 and l ( t 7 ) l ( t 1 ), t 2 = t 4 t 1 1 b 1 t 1 forsome t 4 2h a;b;c i ,so t 1 3 cbat 3 = t 1 1 bt 1 t 1 4 ct 4 a = t 1 1 bt 1 at 1 5 ct 5 ; 74 where t 5 = t 4 a .Notethatthelastletterof t 4 mustbe a 1 ,so l ( t 5 ) l ( t 5 ), t 3 = t 6 t 1 5 ct 5 forsome t 6 2h a;b;c i ,so t 1 5 c 1 t 5 t 1 6 cbat 6 = t 1 1 bt 1 a = ) t 1 6 cbat 6 = t 1 5 ct 5 t 1 1 bt 1 a .Now, t 6 = t 3 t 1 5 c 1 t 5 playstheroleof t 3 ; t 5 = t 2 t 1 1 bt 1 a playstheroleof t 2 ; t 1 doesnotchangeand l ( t 6 ) l ( t 5 ), t 1 1 = t 1 5 bt 5 t 1 6 for t 6 2h a;b;c i ,so t 1 6 b 1 t 6 t 1 5 c 1 t 5 = at 1 3 cbat 3 = ) t 1 3 cbat 3 = t 1 5 ct 5 t 1 6 bt 6 a .Now, t 3 doesnotchange; t 5 = t 2 t 1 3 cbat 3 a 1 playstheroleof t 2 ; t 6 = t 1 t 1 5 bt 5 playstheroleof t 1 and l ( t 5 )